## 15.1 Simple Harmonic Motion

Learning objectives.

By the end of this section, you will be able to:

• Define the terms period and frequency
• List the characteristics of simple harmonic motion
• Explain the concept of phase shift
• Write the equations of motion for the system of a mass and spring undergoing simple harmonic motion
• Describe the motion of a mass oscillating on a vertical spring

When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time ( Figure 15.2 ). The string vibrates around an equilibrium position, and one oscillation is completed when the string starts from the initial position, travels to one of the extreme positions, then to the other extreme position, and returns to its initial position. We define periodic motion to be any motion that repeats itself at regular time intervals, such as exhibited by the guitar string or by a child swinging on a swing. In this section, we study the basic characteristics of oscillations and their mathematical description.

## Period and Frequency in Oscillations

In the absence of friction, the time to complete one oscillation remains constant and is called the period ( T ) . Its units are usually seconds, but may be any convenient unit of time. The word ‘period’ refers to the time for some event whether repetitive or not, but in this chapter, we shall deal primarily in periodic motion, which is by definition repetitive.

A concept closely related to period is the frequency of an event. Frequency ( f ) is defined to be the number of events per unit time. For periodic motion, frequency is the number of oscillations per unit time. The relationship between frequency and period is

The SI unit for frequency is the hertz (Hz) and is defined as one cycle per second :

A cycle is one complete oscillation .

## Example 15.1

Determining the frequency of medical ultrasound.

Solve to find

## Significance

Characteristics of simple harmonic motion.

A very common type of periodic motion is called simple harmonic motion (SHM) . A system that oscillates with SHM is called a simple harmonic oscillator .

## Simple Harmonic Motion

In simple harmonic motion, the acceleration of the system, and therefore the net force, is proportional to the displacement and acts in the opposite direction of the displacement.

A good example of SHM is an object with mass m attached to a spring on a frictionless surface, as shown in Figure 15.3 . The object oscillates around the equilibrium position, and the net force on the object is equal to the force provided by the spring. This force obeys Hooke’s law F s = − k x , F s = − k x , as discussed in a previous chapter.

If the net force can be described by Hooke’s law and there is no damping (slowing down due to friction or other nonconservative forces), then a simple harmonic oscillator oscillates with equal displacement on either side of the equilibrium position, as shown for an object on a spring in Figure 15.3 . The maximum displacement from equilibrium is called the amplitude ( A ) . The units for amplitude and displacement are the same but depend on the type of oscillation. For the object on the spring, the units of amplitude and displacement are meters.

What is so significant about SHM? For one thing, the period T and frequency f of a simple harmonic oscillator are independent of amplitude. The string of a guitar, for example, oscillates with the same frequency whether plucked gently or hard.

Two important factors do affect the period of a simple harmonic oscillator. The period is related to how stiff the system is. A very stiff object has a large force constant ( k ) , which causes the system to have a smaller period. For example, you can adjust a diving board’s stiffness—the stiffer it is, the faster it vibrates, and the shorter its period. Period also depends on the mass of the oscillating system. The more massive the system is, the longer the period. For example, a heavy person on a diving board bounces up and down more slowly than a light one. In fact, the mass m and the force constant k are the only factors that affect the period and frequency of SHM. To derive an equation for the period and the frequency, we must first define and analyze the equations of motion. Note that the force constant is sometimes referred to as the spring constant .

## Equations of SHM

Consider a block attached to a spring on a frictionless table ( Figure 15.4 ). The equilibrium position (the position where the spring is neither stretched nor compressed) is marked as x = 0 x = 0 . At the equilibrium position, the net force is zero.

Work is done on the block to pull it out to a position of x = + A , x = + A , and it is then released from rest. The maximum x -position ( A ) is called the amplitude of the motion. The block begins to oscillate in SHM between x = + A x = + A and x = − A , x = − A , where A is the amplitude of the motion and T is the period of the oscillation. The period is the time for one oscillation. Figure 15.5 shows the motion of the block as it completes one and a half oscillations after release. Figure 15.6 shows a plot of the position of the block versus time. When the position is plotted versus time, it is clear that the data can be modeled by a cosine function with an amplitude A and a period T . The cosine function cos θ cos θ repeats every multiple of 2 π , 2 π , whereas the motion of the block repeats every period T . However, the function cos ( 2 π T t ) cos ( 2 π T t ) repeats every integer multiple of the period. The maximum of the cosine function is one, so it is necessary to multiply the cosine function by the amplitude A .

Recall from the chapter on rotation that the angular frequency equals ω = d θ d t ω = d θ d t . In this case, the period is constant, so the angular frequency is defined as 2 π 2 π divided by the period, ω = 2 π T ω = 2 π T .

The equation for the position as a function of time x ( t ) = A cos ( ω t ) x ( t ) = A cos ( ω t ) is good for modeling data, where the position of the block at the initial time t = 0.00 s t = 0.00 s is at the amplitude A and the initial velocity is zero. Often when taking experimental data, the position of the mass at the initial time t = 0.00 s t = 0.00 s is not equal to the amplitude and the initial velocity is not zero. Consider 10 seconds of data collected by a student in lab, shown in Figure 15.7 .

The data in Figure 15.7 can still be modeled with a periodic function, like a cosine function, but the function is shifted to the right. This shift is known as a phase shift and is usually represented by the Greek letter phi ( ϕ ) ( ϕ ) . The equation of the position as a function of time for a block on a spring becomes

This is the generalized equation for SHM where t is the time measured in seconds, ω ω is the angular frequency with units of inverse seconds, A is the amplitude measured in meters or centimeters, and ϕ ϕ is the phase shift measured in radians ( Figure 15.8 ). It should be noted that because sine and cosine functions differ only by a phase shift, this motion could be modeled using either the cosine or sine function.

The velocity of the mass on a spring, oscillating in SHM, can be found by taking the derivative of the position equation:

Because the sine function oscillates between –1 and +1, the maximum velocity is the amplitude times the angular frequency, v max = A ω v max = A ω . The maximum velocity occurs at the equilibrium position ( x = 0 ) ( x = 0 ) when the mass is moving toward x = + A x = + A . The maximum velocity in the negative direction is attained at the equilibrium position ( x = 0 ) ( x = 0 ) when the mass is moving toward x = − A x = − A and is equal to − v max − v max .

The acceleration of the mass on the spring can be found by taking the time derivative of the velocity:

The maximum acceleration is a max = A ω 2 a max = A ω 2 . The maximum acceleration occurs at the position ( x = − A ) ( x = − A ) , and the acceleration at the position ( x = − A ) ( x = − A ) and is equal to a max a max .

## Summary of Equations of Motion for SHM

In summary, the oscillatory motion of a block on a spring can be modeled with the following equations of motion:

Here, A is the amplitude of the motion, T is the period, ϕ ϕ is the phase shift, and ω = 2 π T = 2 π f ω = 2 π T = 2 π f is the angular frequency of the motion of the block.

## Example 15.2

Determining the equations of motion for a block and a spring.

Work is done on the block, pulling it out to x = + 0.02 m . x = + 0.02 m . The block is released from rest and oscillates between x = + 0.02 m x = + 0.02 m and x = −0.02 m . x = −0.02 m . The period of the motion is 1.57 s. Determine the equations of motion.

All that is left is to fill in the equations of motion:

## The Period and Frequency of a Mass on a Spring

One interesting characteristic of the SHM of an object attached to a spring is that the angular frequency, and therefore the period and frequency of the motion, depend on only the mass and the force constant, and not on other factors such as the amplitude of the motion. We can use the equations of motion and Newton’s second law ( F → net = m a → ) ( F → net = m a → ) to find equations for the angular frequency, frequency, and period.

Consider the block on a spring on a frictionless surface. There are three forces on the mass: the weight, the normal force, and the force due to the spring. The only two forces that act perpendicular to the surface are the weight and the normal force, which have equal magnitudes and opposite directions, and thus sum to zero. The only force that acts parallel to the surface is the force due to the spring, so the net force must be equal to the force of the spring:

Substituting the equations of motion for x and a gives us

Cancelling out like terms and solving for the angular frequency yields

The angular frequency depends only on the force constant and the mass, and not the amplitude. The angular frequency is defined as ω = 2 π / T , ω = 2 π / T , which yields an equation for the period of the motion:

The period also depends only on the mass and the force constant. The greater the mass, the longer the period. The stiffer the spring, the shorter the period. The frequency is

## Vertical Motion and a Horizontal Spring

When a spring is hung vertically and a block is attached and set in motion, the block oscillates in SHM. In this case, there is no normal force, and the net effect of the force of gravity is to change the equilibrium position. Consider Figure 15.9 . Two forces act on the block: the weight and the force of the spring. The weight is constant and the force of the spring changes as the length of the spring changes.

When the block reaches the equilibrium position, as seen in Figure 15.9 , the force of the spring equals the weight of the block, F net = F s − m g = 0 F net = F s − m g = 0 , where

From the figure, the change in the position is Δ y = y 0 − y 1 Δ y = y 0 − y 1 and since − k ( − Δ y ) = m g − k ( − Δ y ) = m g , we have

If the block is displaced and released, it will oscillate around the new equilibrium position. As shown in Figure 15.10 , if the position of the block is recorded as a function of time, the recording is a periodic function.

If the block is displaced to a position y , the net force becomes F net = k ( y 0 − y ) − m g = 0 F net = k ( y 0 − y ) − m g = 0 . But we found that at the equilibrium position, m g = k Δ y = k y 0 − k y 1 m g = k Δ y = k y 0 − k y 1 . Substituting for the weight in the equation yields

Recall that y 1 y 1 is just the equilibrium position and any position can be set to be the point y = 0.00 m . y = 0.00 m . So let’s set y 1 y 1 to y = 0.00 m . y = 0.00 m . The net force then becomes

This is just what we found previously for a horizontally sliding mass on a spring. The constant force of gravity only served to shift the equilibrium location of the mass. Therefore, the solution should be the same form as for a block on a horizontal spring, y ( t ) = A cos ( ω t + ϕ ) . y ( t ) = A cos ( ω t + ϕ ) . The equations for the velocity and the acceleration also have the same form as for the horizontal case. Note that the inclusion of the phase shift means that the motion can actually be modeled using either a cosine or a sine function, since these two functions only differ by a phase shift.

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## Simple Harmonic Motion Problems for High Schools

Simple harmonic motion is defined as a kind of motion in which the net force along the motion obeys Hooke's law.

According to Hooke's law, the net force is proportional to the displacement from the equilibrium point and is always directed toward that point. $F_{net}=-kx$ This article teaches all concepts relating to simple harmonic motion (SHM) by problem-solution strategy.

In the following, some problems on simple harmonic motion are solved. All problems are for the AP Physics 1 exam and/or high school students.

## Simple Harmonic Motion (SHM) Problems

Problem (1): A 0.50-kg object is attached to a horizontal spring whose spring constant is k=300 N/m and is undergoing a simple harmonic motion. Calculate its (a) Period,    (b) frequency,        (c) angular frequency.

Solution : Imagine an object is attached to an unstretched spring, displaces the spring from its equilibrium, and then releases. After releasing, the object undergoes a simple harmonic motion. The period and frequency of such simple harmonic motion are obtained by the following formula $T=2\pi\sqrt{\frac{m}{k}}\quad,\quad f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ where $m$ is the mass of the object and $k$, is a constant that indicates the stiffness of the spring, called the spring constant.

(a) Substituting the values into the above formula for the period, we have $T=2\pi\sqrt{\frac{0.5}{300}}=0.256\quad {\rm s}$  (b) To find the frequency, we can use the formula above or use $f=1/T$, instead. Thus, $f=\frac{1}{0.256}=3.9\quad {\rm Hz}$ Note that the SI unit of frequency is Hz.

(c) Angular frequency is related to the frequency or period of oscillation as $\omega=2\pi f=2\pi/T$. Therefore, $\omega=2\pi f=2\pi(3.9)=24.5\,{\rm \frac{rad}{s}}$ The SI unit of angular frequency is radians per second, $rad/s$.

Problem (2): A scale is stretched by 3.2 cm when a 2.4 kg object is attached.  (a) Find the spring stiffness constant. (b) Now suppose the object is lowered downward from its rest by as much as 2 cm and released. Find the period and frequency of the oscillations.  (c) Write down the vertical displacement vs. time equation for this system.

Solution : The spring-mass system has a simple harmonic motion in which the period and frequency of oscillations are given by the following formula $T=2\pi\sqrt{\frac{m}{k}}\quad,\quad f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ where $k$ is the spring stiffness constant.

(a) In this problem, $k$ is unknown and is found by applying Newton's second law of motion to the vertical forces, downward weight, and upward spring force, acting on the object. These forces are in balance because the object is at rest after stretching.

Thus, when a load is hung from a spring (say a scale), the spring constant is computed as below: [k=\frac{mg}{x_0}=\frac{(2.4)(9.8)}{3.2}=7.35\,({\rm N \cdot m^{-1}})\] In the above, $x_0$ is the maximum distance when a load stretches an unloaded spring.   (b) The initial configuration (after hanging an object from the scale) is at rest with no oscillations. This situation is called the equilibrium state.

Any agent that disrupts this situation can cause oscillations in the mass-spring system.

In this example problem, the object is pushed from its equilibrium point by $2\,{\rm cm}$ and comes to another rest. This maximum displacement is called the amplitude of the oscillations. Thus, $A=0.2\,{\rm m}$.

The period, which is the time required to complete one cycle, is obtained as below $T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{2.4}{7.35}}=3.6\,{\rm s}$ The frequency of the oscillations, which indicates the number of full cycles per second is the inverse of the period $f=\frac{1}{T}=\frac{1}{3.6}=0.28\,{\rm Hz}$  (c) The vertical displacement vs. time equation for a simple harmonic motion is as $y=A\sin(\omega t)$ or $y=A\cos(\omega)t$ which depends on the initial configuration of the system. $\omega=2\pi/T=2\pi f$ is the angular frequency.

The former case is for when the object is attached to an unstretched spring, i.e., $x_0=0$. The latter case is for an object that is hung from a stretched spring, i.e., $x_0\neq0$, see figures.

This spring-mass system starts its oscillations when the spring is initially stretched some length, so we must use the cosine function. \begin{align*} y&=A\cos\left(\frac{2\pi}{T} t\right)\\\\&= (0.2)\cos\left(\frac{2\pi}{3.6} t\right)\\\\&=(0.2\,{\rm m})\cos\left(\frac{5\pi}{9} t\right)\end{align*}

Problem (3): An object of 45 N is hanging from a spring vertically. The spring is stretched from its equilibrium point by 0.14 m. What is the spring constant?

Solution : The spring is initially unloaded and unstretched. Hanging a 45-N object from it causes it to stretch by 0.14 m. This system is at rest and, therefore, at equilibrium. Applying Newton's second law gives the following formula to find the spring stiffness constant $k=\frac{mg}{x_0}=\frac{45}{0.14}=321.4 \, {\rm N\cdot m^{-1}}$ Note that $x_0$ is the length the unloaded spring is initially stretched by an object's weight.

Problem (4): A 2-kg block is attached to a spring whose constant is 32 N/m horizontally. Imagine it being displaced from equilibrium by 4 cm and released from rest. How much time does it take for the block to reach the point x=2 cm?

Solution : To find the required time, we must write down an equation in which the object's position at any instant of time is shown explicitly.

We know that a mass attached to a spring undergoes a simple harmonic motion (SHM) with the equation $x=A\cos\omega t$.

Above, $A$ is the maximum displacement from the equilibrium point of the spring-mass system, called the amplitude.

$\omega$ is also the angular frequency, which is related to period and frequency by $\omega=2\pi f=2\pi/T$.

The block's period of oscillation is found by $T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{2}{32}}=\frac{\pi}{4}\,{\rm s}$ Therefore, the angular frequency is $\omega=2\pi/T=8\,{\rm rad/s}$.

In this example, the maximum distance (amplitude) from where the block is released at time $t=0$ is 4 cm, so $A=0.4\,{\rm m}$.

Substituting these quantities into the above equation of SHM yields $x=(0.4\,{\rm m})\cos(8t)$. This equation gives the location of the object at any instant of time.

If we put $x=0.2\,{\rm m}$ into the equation above and solve for the time $t$, it gives the time required to reach that point. \begin{align*} x&=0.4 \cos(8t)\\ 0.2&=0.4 \cos(8t)\\ \Rightarrow 1/2&=\cos(8t)\end{align*} Taking inverse trigonometric of both sides, gives $8t=\cos^{-1}(1/2)=\pi/6 \Rightarrow t=\pi/48=0.06\,{\rm s}$ Hence, it takes 0.06 seconds for the object to reach the point located at x=2 cm.

Problem (5): An object hangs from a spring with a constant stiffness of 40 N/m and stretches it by 0.5 m.  (a) What is the mass of the hanging object? (take $g=10\,{\rm m/s^2}$) (b) We pull the object $3\,{\rm cm}$ from its equilibrium position and release it. Find an equation for the position of the object as a function of time.

Solution : (a) Because the object is at rest, applying Newton's second law gives the equation $kx=mg$. Solving for the unknown mass $m$, we have $m=\frac{kx}{g}=\frac{40\times 0.5}{10}=2\,{\rm kg}$ (b) By doing this, the object oscillates around its equilibrium position. Here, $0.3\,{\rm m}$ is the maximum distance from that position, which is called the amplitude of oscillations.

The period of the oscillations is also obtained by the following formula: \begin{align*} T&=2\pi\sqrt{\frac{m}{k}}\\\\&=2\pi\sqrt{\frac{2}{40}}\\\\&=1.4\quad {\rm s}\end{align*} Thus, one complete cycle takes 2 seconds.

When an object at the maximum distance from its equilibrium starts its oscillation motion, then the standard equation of position vs. time is given by $y=A\cos(\omega t)$, where $\omega=2\pi/T$.

Therefore, the displacement vs. time equation for this object is as follows: $y=(0.3\,{\rm m})\cos\left(\frac{10\pi}{7} t\right)$

Problem (6): Consider placing a spring in a vertical position and putting a 100-g object on it. We see it compressed by 2 cm.  (a) If an additional load of 300 grams is placed on top of the previous object, how much will the spring compress this time? (b) Find the spring constant.

Solution : This question is similar to the problems on Hooke's law , refer to that page and practice more. We must use Hooke's law twice. Here, the weight of the object, $W$, causes compression. The first object brings the spring to an equilibrium position, so $k\Delta x_1=W_1$.

The additional load also makes the displacement $k\Delta x_2=W_2$. Taking the proportions of these two equations and solving for the unknown displacement $\Delta x_2$, we have \begin{align*} \frac{k\Delta x_1}{k\Delta x_2}&=\frac{W_1}{W_2} \\ \\\Rightarrow \Delta x_2&=\frac{W_2 \Delta x_1}{W_1}\\\\ &=\frac{300\times 2}{100}\\\\&=6\quad {\rm cm}\end{align*} Therefore, the second object compresses the spring 6 cm from its previous position.

(b) The weight of the first load $(0.1\,{\rm kg})(9.8)=0.98\,{\rm N}$ causes the spring to compress by $0.2\,{\rm m}$. By applying Hooke's law, we can find the spring constant as below $k=\frac{F}{x}=\frac{0.98}{0.2}=4.9\,({\rm N\cdot m^{-1}})$

Problem (7): An object attached to a spring with a constant of 400 N/m vibrates at 36 Hz. If the stiffness of the spring is doubled, how would the oscillations change?

Solution : According to the frequency formula for a simple harmonic motion, $f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$, the frequency of vibrations is proportional to the square root of the spring's constant, i.e., $f\propto \sqrt{k}$. Thus, we would have \begin{align*} \frac{f_2}{f_1}&=\frac{\sqrt{k_2}}{\sqrt{k_1}}\\\\\frac{f_2}{f_1}&=\frac{\sqrt{2\times 400}}{\sqrt{400}}\\\\ &=\sqrt{2} \end{align*} Hence, the frequency of vibrations becomes $\sqrt{2}$ times the frequency of the original spring.

Problem (8): A cork is on the surface of rippling water and does have an up-and-down motion. The cork's vertical displacement motion $y$ is described by the following function $y(t)=3.3 \sin\left(\frac 13 t+\frac{\pi}5\right)$ All quantities are in SI units. (a) Find the amplitude, frequency, period, angular frequency, and phase constant of the motion.  (b) Where is the cork at time $t=2\,{\rm s}$.  (c) Find the velocity and acceleration of the motion as functions of time and their corresponding maximum values. (d) Find the initial values of the position, velocity, and acceleration of the cork.

Solution : The standard equation of motion for simple harmonic motion is given by the formula $y(t)=A\sin(\omega t+\delta)$ where $A$ is the amplitude, $\omega$ is the angular frequency, and $\delta$ is the phase constant. The angular frequency is also related to the period $T$ and frequency $f$ as $\omega=2\pi f=2\pi/T$.

(a) In this problem, the given function is similar to the function of simple harmonic motion (SHM). By comparing these two functions, we can find the quantities asked for. Therefore, \begin{gather*} A=3.3\quad{\rm m} \\ \omega=\frac 13\quad{\rm rad/s} \\ T=\frac{2\pi}{\omega}=6\pi\quad{\rm s}\\ f=\frac{\omega}{2\pi}=\frac{2}{3\pi}\quad{\rm Hz}\\ \delta=\pi/5 \quad {\rm rad}\end{gather*} (b) Setting $t=2\,{\rm s}$ gives the vertical position of the cork at any instant of time relative to the water level. $y(t=2)=3.3 \sin\left(\frac 13 (2)+\frac{\pi}5\right)=3.17\,{\rm m}$ (c) Velocity is the first derivative of the position function, so we have \begin{align*} v_y&=\frac{dy}{dt}=\frac{d}{dt} A\sin(\omega t+\delta)\\\\&=A\omega \cos(\omega t+\delta)\\\\&=(3.3)(1/3)\cos \left(\frac 13 t+\frac{\pi}5\right)\\\\&=(1.1\,{\rm m/s}) \cos \left(\frac 13 t+\frac{\pi}5\right)\end{align*} The acceleration is also obtained by differentiation with respect to velocity or second derivative of the position. \begin{align*} a_y&=\frac{dv_y}{dt}=\frac{d}{dt} A\omega\cos(\omega t+\delta)\\\\&=-A\omega^2 \sin(\omega t+\delta)\\\\&=-(3.3)(1/3)^2 \sin \left(\frac 13 t+\frac{\pi}5\right)\\\\&=-(0.37\,{\rm m/s^2}) \sin \left(\frac 13 t+\frac{\pi}5\right)\end{align*} Recall that the maximum value of sine or cosine functions like $A\cos(anything)$ or $A\sin(anything)$ is $A$. Therefore, the maximum values of velocity and acceleration are found as below $v_{max}=1.1\,{\rm m/s} \quad,\quad a_{max}= 0.37\,{\rm m/s^2}$ (d) By setting $t=0$ in the vertical displacement function $y(t)$, we can find the asked initial values. \begin{align*} y_0(t=0)&=(3.3)\sin(1/3\times 0+\pi/5)\\&=(3.3)\sin(\pi/5)\\&=1.94\,{\rm m}\\\\v_{0y}(t=0)&=(1.1)\cos(1/3\times 0+\pi/5)\\&=0.9\,{\rm m/s}\\\\a_{0y}(t=0)&=-(0.37)\sin(1/3\times 0+\pi/5)\\&=-0.22\,{\rm m/s^2}\end{align*}

Problem (9): The horizontal displacement of an object attached to a spring is described by the following equation $x=0.5\cos (0.4t)$ where $x$ is in meters and $t$ is in seconds. Find the (a) amplitude, (b) frequency, (c) period, (d) maximum velocity, and (e) maximum acceleration.

Solution : The displacement of this object is varying as a cosine function, which is a characteristic feature of a simple harmonic motion with the standard equation $x=A\sin(\omega t)$ or $x=A\cos(\omega t)$. The angular frequency is also defined as $\omega=2\pi/T=2\pi f$ where $T$ and $f$ are the period and frequency of oscillations, respectively. By comparing the given equation with the standard form, one can find the requested quantities.

(a) The amplitude, which is defined as the maximum distance from the equilibrium position in an oscillating motion, is found as $A=0.5\,{\rm m}$.

(b) The period of the harmonic motion is the time it takes the motion to repeat itself and is obtained from the angular frequency by the following formula $T=\frac{2\pi}{\omega}=\frac{2\pi}{0.4}=5\pi\,{\rm s}$ (c) The frequency of harmonic motion, which is defined as the number of complete cycles in one second, is related to the angular frequency as below $f=\frac{\omega}{2\pi}=\frac{0.4}{2\pi}=\frac{1}{5\pi}\,{\rm Hz}$ Keep in mind that the period and frequency are also related together by $f=1/T$.

(d) The maximum velocity in a simple harmonic motion is found as $v_{max}=A\omega$, thus in this case, we have $v_{max}=(0.5)(0.4)=0.2\,{\rm m/s}$  (e) The maximum acceleration in a simple harmonic motion is also found by the formula $a_{max}=A\omega^2$, therefore, $a_{max}=(0.5)(0.4)^2=0.08\,{\rm m/s^2}$

Problem (10): An object of mass 0.5 kg attached to a massless spring with spring constant k=15 N/m. It is displaced from its equilibrium by 4 cm and released.

(a) Find the total energy of the mass-spring system.      Solution : Recall that the total mechanical energy of a system consisting of a mass attached to a spring remains constant during back-and-forth oscillations.

The value of this constant is found by the formula $E=\frac 12 kA^2$, where $A$ is the amplitude of the spring.

In this example, the greatest distance from equilibrium (amplitude) is 5 cm so $A=0.4\,{\rm m}$. Therefore, the total mechanical energy of this system is $E=\frac 12 kA^2=\frac 12 (15)(0.04)^2= 0.012\,{\rm J}$  (b) Compute the elastic potential energy of the spring when it is located $2\,{\rm cm}$ from the releasing point.

Solution : When a spring is stretched or compressed, it stores a type of potential energy called elastic potential energy, whose formula is $U_e=\frac{1}{2}kx^2$ Where $x$ is the amount of stretching or compression from its equilibrium point.

Here, the total displacement from the equilibrium point is $\Delta x=4-2=2\,{\rm cm}$. Thus, the elastic potential energy of spring stored at that location is $U_e=\frac{1}{2} (15\,{\rm N/m})(0.02\,{\rm m})^2=0.003\,{\rm J}$

(c) Compute the maximum velocity of the object.

We know that the sum of elastic potential energy $U_e$ and kinetic energy is the total mechanical energy of a mass-spring system, which is a constant value.

On the other hand, when $U_e$ is zero, the kinetic energy acquires its maximum value, so the mechanical energy remains constant. Therefore, at that point, we have \begin{align*} E&=P_e+\frac 12 mv^2\\\\0.012&=0+\frac 12 (0.5)v_{max}^2\\\\\Rightarrow v_{max}&=\sqrt{0.048}\\\\&=0.22\quad {\rm m/s}\end{align*}

(d) What is the velocity of the object when the block is placed 2 cm away from the spring?

Again, using the conservation of mechanical energy in the mass-spring system, we can find the velocity at each point of the path of the oscillation. \begin{align*} E&=P.E+K.E\\\\ E&=\frac 12 kx^2 + \frac 12 mv^2\\\\ 0.012 &=\frac 12 (15)(0.02)^2+ \frac 12 (0.5)v^2 \\\\\Rightarrow v&=\sqrt{3.6}=0.19\,{\rm m/s}\end{align*}

In the following article, you can find the exact meaning of potential energy in physics with some solved problems .

Problem (11): A block of mass 2 kg is attached to a spring with a spring constant of 500 N/m that lies horizontally on a surface. The block is pulled to a point 5 cm away from the equilibrium point. Calculate  (a) The work required to stretch the spring. (b) the speed of the block when it passes through equilibrium.

Solution : (a) The work done for stretching or compressing a spring is stored in the form of potential energy, called elastic potential energy, in the spring.

The formula for elastic potential energy is as $U_s=\frac 12 kx^2$, where $x$ is the distance from the equilibrium point.

The equilibrium point is also defined as the point where neither the spring is stretched nor compressed. Therefore, $U_s=\frac 12 kx^2=\frac 12 (500)(0.05)^2=0.0625\,{\rm J}$ The work done to bring the spring 5 cm to the right of equilibrium is about 0.06 joules.

(b) The conservation of mechanical energy in the case of the spring-mass system has the following form $\frac 12 kA^2=\frac 12 kx^2+\frac 12 mv^2$ where $A$ is the maximum displacement from the equilibrium (amplitude), $x$ is the position of the object attached to the spring at any instant of time relative to the equilibrium, and $v$ is the object's velocity at that specific point.

In this problem, the velocity is asked when the object is in equilibrium, that is, when $x=0$. On the other hand, the spring initially is stretched by $5\,{\rm cm}$ and released from rest, so this value is the amplitude of the oscillations. Therefore, \begin{align*} \frac 12 kA^2&=\frac 12 kx^2+\frac 12 mv^2 \\ \\ \frac 12 (500)(0.05)^2&=\frac 12 k (0)+\frac 12 (2)v^2 \\ \\ \Rightarrow v&=0.8\quad {\rm m/s}\end{align*} Hence, when the block passing through the equilibrium position, its velocity is about $0.8\,{\rm m/s}$.

It is worth noting that, at this position, the speed of the object attached to the spring is always at its maximum.

(c) Suppose the coefficient of friction between the block and horizontal surface is $\mu_k=0.250$. Again, find the above speed at the equilibrium point.

Author : Dr. Ali Nemati Date Published: 7/18/2021

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## 13.7: Sample problems and solutions

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## Exercise $$\PageIndex{1}$$

Ty ( $$m=30\text{kg}$$ ) is trying out a new piece of equipment at his local playground. The equipment consists of a platform that is connected to two springs. The top spring ( $$k_1=2400\text{N/m}$$ ) connects the platform to the playground structure and the bottom spring ( $$k_2=3480\text{N/m}$$ ) (Figure $$\PageIndex{1}$$) connects it to the ground. When no one is standing on the platform the platform is $$50\text{cm}$$ off the ground. When Ty is standing on the platform, he oscillates up and down, and the lowest point that the platform reaches is $$35\text{cm}$$ off the ground. Show that this is simple harmonic motion and determine what Ty’s maximum speed will be.

First, we need to solve for the new equilibrium position of the platform, $$x_0$$ , when Ty is standing on the platform. We define the $$x$$ axis so that the origin is $$50\text{cm}$$ above the ground (the equilibrium position when no one is standing on the platform) and choose the positive direction to be downwards (Figure $$\PageIndex{2}$$).

Even though we do not know the mass of the platform, or the actual resting lengths of the spring, we do not need to know these, since we can model the platform with nobody on it as a single spring with spring constant $$k=k_1+k_2$$ and rest position $$x=0$$ .

When Ty is standing on the platform, the sum of the forces is given by his weight and the force from the “effective spring”:

\begin{aligned} \sum F=mg-(k_1+k_2)x\end{aligned}

where we noted that, when the platform moves down, both the top and bottom spring will exert a force upwards (Figure $$\PageIndex{3}$$).

At equilibrium, the sum of the forces is equal to zero. We can use this to solve for the displacement at $$x_0$$ :

\begin{aligned} 0&=mg-(k_1+k_2)x_0\\[4pt] \therefore x_0&=\frac{mg}{k_1+k_2}=\frac{(30\text{kg})(9.8\text{m/s}^{2})}{(2400\text{Nm})+(3480\text{Nm})}=0.05\text{m}\end{aligned}

We will confirm that this is a simple harmonic oscillator by showing that the system’s motion can be described by the equation:

\begin{aligned} \frac{d^2x}{dt^2}&=-\omega^2x\end{aligned}

For some position $$x$$ below equilibrium, we can rewrite Newton’s second law as:

\begin{aligned} ma&=mg-(k_1+k_2)x\\[4pt] m\frac{d^2x}{dt^2}&=mg-(k_1+k_2)x\end{aligned}

In order to show that this is simple harmonic motion, we need to combine the right hand side of the equation into one term. We found earlier that $$mg=(k_1+k_2)x_0$$ , which we can use here:

\begin{aligned} m\frac{d^2x}{dt^2}&=(k_1+k_2)x_0-(k_1+k_2)x\\[4pt] \frac{d^2x}{dt^2}&=\frac{(k_1+k_2)}{m}(x_0-x)\\[4pt] \frac{d^2x}{dt^2}&=-\frac{(k_1+k_2)}{m}(x-x_0)\\[4pt]\end{aligned}

We now define an $$x'$$ axis such that $$x'=x-x_0$$ . This means that the origin of the $$x'$$ axis is at the new equilibrium position:

We can now rewrite our expression using the $$x'$$ axis:

\begin{aligned} \frac{d^2x}{dt^2}&=-\frac{(k_1+k_2)}{m}x'\end{aligned}

This equation tells us that this is simple harmonic motion about the new equilibrium position, where $$\omega=\sqrt{(k_1+k_2)/m}$$ . We know that the lowest point that the platform reaches is 35 cm above the ground, which, on our $$x'$$ axis, corresponds to $$x'=10\text{cm}$$ (Figure $$\PageIndex{3}$$). Thus, the amplitude of the oscillation is $$A=0.1\text{m}$$ . Because this is simple harmonic motion, we know that the position of the platform can be described by the following function:

\begin{aligned} x'(t)&= A \cos(\omega t + \phi)\end{aligned}

We set $$t=0$$ to be when the platform is at its lowest point ( $$x'=A$$ ). The value of $$\phi$$ is thus:

\begin{aligned} x'(0)&= A \cos(\omega (0) + \phi)\\[4pt] A&= A \cos(\phi)\\[4pt] 1&=\cos(\phi)\\[4pt] \therefore \phi&=0\end{aligned}

The velocity is given by:

\begin{aligned} v(t)=\frac{d}{dt}x(t) &= -A\omega\sin(\omega t + \phi)\\[4pt] &=-A\omega\sin(\omega t)\end{aligned}

The speed will be maximized when $$\sin(\omega t)=1\quad \textrm{or} -1$$ . So, the maximum speed will be:

\begin{aligned} |v|&=A\omega\\[4pt] |v|&=A\sqrt{\frac{(k_1+k_2)}{m}}\\[4pt] |v|&=(0.1\text{m})\sqrt{\frac{(2400\text{Nm}+3480\text{Nm})}{30\text{kg}}}\\[4pt] |v|&=1.4\text{m/s}\end{aligned}

## Exercise $$\PageIndex{2}$$

A torsional pendulum consists of a horizontal rod suspended from a vertical wire. When the rod is rotated so that it is displaced an angle $$\theta$$ from equilibrium, the wire (which is now twisted) provides a restoring torque about the axis of the wire given by:

\begin{aligned} \tau=-\kappa\theta\end{aligned}

where $$\kappa$$ is the torsion coefficient, which depends on the stiffness of the wire. You may notice that this formula closely resembles Hooke’s law.

a. You construct a torsional pendulum by attaching two small spherical masses (you can assume they are point masses, each of mass $$m$$ ) to the ends of a thin (mass-less) rod of length $$L$$ and attaching a wire to the center of the rod (Figure $$\PageIndex{4}$$). When you displace one of the masses by an angle $$\theta$$ and release it, you find that it oscillates with a period $$T$$ . Find an expression for the torsion coefficient, $$\kappa$$ , in term of $$T$$ , $$m$$ , and $$L$$ .

b. You place two very large spheres, each of mass $$M$$ , near each of the small spheres (as shown in Figure $$\PageIndex{5}$$). Each of the small spheres will be acted on by a force of gravity from the nearest large sphere. The pendulum is at equilibrium when it is deflected an angle $$\beta$$ from its original equilibrium position. At the new equilibrium, the displacement vectors connecting the centers of large and small spheres have a magnitude $$d$$ and are essentially perpendicular to the rod. Find an expression for the universal gravitational constant $$G$$ , in terms of the masses, the length of the rod, and the period measured in part a).

Fun fact! This set-up resembles an experiment performed by Henry Cavendish that was first used to determine the value for $$G$$ and to test Newton’s Universal Theory of Gravity.

The only force that creates a torque on the masses is the restoring force from the twisting of the wire. The rotational dynamics version of Newton’s Second Law relates this torque to the angular acceleration, $$\alpha$$ of the rod:

\begin{aligned} I\alpha=-\kappa\theta\end{aligned}

where $$I$$ is the moment of inertia of the rod. Rewriting $$\alpha$$ more explicitly as the second time derivative of the angle, we get:

\begin{aligned} I\frac{d^2\theta}{dt^2}&=-\kappa\theta\\[4pt] \frac{d^2\theta}{dt^2}&=-\frac{\kappa}{I}\theta\\[4pt]\end{aligned}

By inspection, we can see that the torsional pendulum is a simple harmonic oscillator, where $$\omega=\sqrt{\kappa/I}$$ . The period of the motion is therefore:

\begin{aligned} T&=\frac{2\pi}{\omega}\\[4pt] T&=2\pi\sqrt{\frac{I}{\kappa}}\end{aligned}

We can rearrange this expression to get $$\kappa$$ :

\begin{aligned} T^2&=\frac{4\pi^2I}{\kappa}\\[4pt] \kappa&=\frac{4\pi^2I}{T^2}\end{aligned}

The moment of inertia for one of the masses is $$m(L/2)^2$$ , where $$L/2$$ is the distance from the mass to the axis of rotation. The moment of inertia for the two masses attached to the mass-less rod is:

\begin{aligned} I&=2m\left(\frac{L}{2}\right)^2=\frac{mL^2}{2}\\[4pt]\end{aligned}

Putting this into our expression for $$\kappa$$ :

\begin{aligned} \kappa=\frac{2\pi^2mL^2}{T^2}\end{aligned}

The two forces that provide torques for the small spheres are gravity and the force exerted by the twisting wire. Each of the small spheres will experience a force due to gravity from the nearest large sphere. At equilibrium, the force due to gravity on one of the small spheres is therefore:

\begin{aligned} F_g=\frac{GMm}{d^2}\end{aligned}

Assuming that, at equilibrium, the force vector is perpendicular to the rod, the torque from one of the large spheres is just the force multiplied by the distance to the axis of rotation. Since there are two large spheres, each of which creates a torque on the pendulum, the total torque due to gravity is:

\begin{aligned} \tau_g&=2F_g\frac{L}{2}\\[4pt] &=F_gL\\[4pt] &=\frac{GMm}{d^2}L\end{aligned}

(Note that $$\tau g$$ is the torque due to gravity at equilibrium only ). We can use Newton’s second law for the pendulum to find an expression for $$G$$ . At equilibrium, the net torque is equal to zero, and the angle of deflection is $$\beta$$ :

\begin{aligned} \tau_{net}&=\tau_{wire}-\tau_g\\[4pt] 0&=\tau_{wire}-\tau_g\\[4pt] \tau_g&=\tau_{wire}\\[4pt] \frac{GMm}{d^2}L&=\kappa\beta\\[4pt] \therefore G&=\frac{\kappa\beta d^2}{LMm}\end{aligned}

Using our expression for $$\kappa$$ found in part a), this becomes:

\begin{aligned} G=\frac{2\pi^2L\beta d^2}{MT^2}\end{aligned}

## 47 Simple Harmonic Motion: General Solution

Exercise 47.1: Another Solution

C. Let’s drive home the point of this exercise with one last question.

The big take away here is this:  the initial conditions of the problem tell you which solution you should use. This makes a lot of sense: if an object is oscillating back and forth, whether the oscillation is described by a sine or a cosine depends on when you “start” looking at the oscillation.  Moreover, recall that

Exercise 47.2: Spring Collision

A. What is the velocity of the mass when the spring is fully compressed?

B.  Find how much the spring is compressed.

I want to walk you through the solution.  We start by describing what happens in English.

Now draw a picture.  The story has a beginning and end: draw both, and label everything.

Key Takeaways

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## Unit 8: Simple harmonic motion

Remember swingsets? You can swing high on them, but you can't get the swing to do a full circle. This is because of harmonic motion, which keeps an object oscillating (moving back and forth) within a specific range of motion. Let's explore harmonic motion and its applications.

## Introduction to simple harmonic motion

• Intuition about simple harmonic oscillators (Opens a modal)
• Definition of amplitude and period (Opens a modal)
• Equation for simple harmonic oscillators (Opens a modal)
• Introduction to simple harmonic motion review (Opens a modal)
• Simple harmonic motion: Finding frequency and period from graphs Get 3 of 4 questions to level up!
• Simple harmonic motion: Finding speed, velocity, and displacement from graphs Get 3 of 4 questions to level up!

## Simple harmonic motion in spring-mass systems

• Period dependence for mass on spring (Opens a modal)
• Simple harmonic motion in spring-mass systems review (Opens a modal)
• Spring-mass systems: Calculating frequency, period, mass, and spring constant Get 3 of 4 questions to level up!
• Analyzing graphs of spring-mass systems Get 3 of 4 questions to level up!

## Simple pendulums

• Pendulums (Opens a modal)
• Simple pendulum review (Opens a modal)
• Period and frequency of simple pendulums Get 3 of 4 questions to level up!

## Energy in simple harmonic oscillators

• Energy graphs for simple harmonic motion (Opens a modal)
• Energy of simple harmonic oscillator review (Opens a modal)
• Analyzing energy for a simple harmonic oscillator from graphs Get 3 of 4 questions to level up!
• Analyzing energy for a simple harmonic oscillator from data tables Get 3 of 4 questions to level up!

## Simple harmonic motion

Simple harmonic motion (SHM) is an mechanical example of periodic motion, a motion of an object that regularly repeat itself. If a force is directed toward the equilibrium position, that motion is referred as harmonic motion. Generally, this force directed toward the eqilibrium position, the opposite of t he displacement vector, is often called a restoring force. The most common form of restoring force is the Hooke's Law. Percisely, a system demonstrates simple harmonic motion when an object's accerleration is porportional to its position and is oppositely directed to the displacement from equilibrium position.

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## Oscillations | Physics - Solved Example Problems for Simple Harmonic Motion (SHM) | 11th Physics : UNIT 10 : Oscillations

Chapter: 11th physics : unit 10 : oscillations, solved example problems for simple harmonic motion (shm), displacement, velocity, acceleration and its graphical representation – shm, example 10.3.

Which of the following represent simple harmonic motion?

(i)  x  =  A  sin ω t  + B cos  ωt

(ii)  x  =  A  sin ωt+ B cos 2 ωt

(iii)  x  =  A e iωt

(iv)  x  =  A  ln  ωt

(i)  x  =  A  sin ω t  +  B  cos  ωt

This differential equation is similar to the differential equation of SHM (equation 10.10).

Therefore,  x  =  A  sin  ωt + B  cos  ωt  represents SHM.

(ii)  x  =A sin  ωt  +  B  cos2 ωt

This differential equation is not like the differential equation of a SHM (equation 10.10). Therefore,  x  =  A  sin  ωt  +  B  cos 2 ωt  does not represent SHM.

(iii) x=Ae j ω t

This differential equation is like the differential equation of SHM (equation 10.10). Therefore,  x  =  A e iωt represents SHM.

(iv)  x  =  A ln  ω t

This differential equation is not like the differential equation of a SHM (equation 10.10). Therefore,  x  =  A  ln ω t  does not represent SHM.

## EXAMPLE 10.4

Consider a particle undergoing simple harmonic motion. The velocity of the particle at position  x 1  is  v 1  and velocity of the particle at position  x 2  is  v 2 . Show that the ratio of time period and amplitude is

## EXAMPLE 10.5

A nurse measured the average heart beats of a patient and reported to the doctor in terms of time period as 0.8 s . Express the heart beat of the patient in terms of number of beats measured per minute.

Let the number of heart beats measured be  f . Since the time period is inversely proportional to the heart beat, then

## EXAMPLE 10.6

Calculate the amplitude, angular frequency, frequency, time period and initial phase for the simple harmonic oscillation given below

a. y  = 0.3 sin (40πt + 1.1)

b.  y  = 2 cos (πt)

c.  y  = 3 sin (2πt − 1.5)

Simple harmonic oscillation equation is y = A sin(ωt + φ 0 ) or y =A cos(ωt + φ 0 )

## EXAMPLE 10.7

Show that for a simple harmonic motion, the phase difference between

a. displacement and velocity is π/2 radian or 90°.

b. velocity and acceleration is π/2 radian or 90°.

c.  displacement and acceleration is π radian or 180°.

a.    The displacement of the particle executing simple harmonic motion

y  =  A  sinω t

Velocity of the particle is

v  = Aωcos ωt = Aωsin(ωt+ π /2)

The phase difference between displacement and velocity is π/2.

b. The velocity of the particle is

v = A ω cos ωt

Acceleration of the particle is

a  = Aω 2 sin ω t = Aω 2 cos(ωt+ π /2)

The phase difference between velocity and acceleration is π/2.

c. The displacement of the particle is  y  =  A  sinω t

a  = −  A  ω 2   sin ω t  = A ω 2   sin(ω t  + π)

The phase difference between displacement and acceleration is π.

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## Chapter 2: Vectors and Scalars

Chapter 3: motion along a straight line, chapter 4: motion in two or three dimensions, chapter 5: newton's laws of motion, chapter 6: application of newton's laws of motion, chapter 7: work and kinetic energy, chapter 8: potential energy and energy conservation, chapter 9: linear momentum, impulse and collisions, chapter 10: rotation and rigid bodies, chapter 11: dynamics of rotational motions, chapter 12: equilibrium and elasticity, chapter 13: fluid mechanics, chapter 14: gravitation, chapter 15: oscillations, chapter 16: waves, chapter 17: sound, chapter 18: temperature and heat, chapter 19: the kinetic theory of gases, chapter 20: the first law of thermodynamics, chapter 21: the second law of thermodynamics, chapter 22: electric charges and fields, chapter 23: gauss's law, chapter 24: electric potential, chapter 25: capacitance, chapter 26: current and resistance, chapter 27: direct-current circuits, chapter 28: magnetic forces and fields, chapter 29: sources of magnetic fields, chapter 30: electromagnetic induction, chapter 31: inductance, chapter 32: alternating-current circuits, chapter 33: electromagnetic waves.

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Consider a car moving over a bumpy road. Here, the suspension of the car acts like a spring connecting the mass of the car to the wheels.

During its motion, the spring compresses and expands, executing simple harmonic motion.

Suppose the car bounces with a certain amplitude; what is its maximum velocity, and how does its total energy change if it is halfway between its initial and equilibrium positions?

The total energy in simple harmonic motion remains conserved. Rearrangement of terms gives the velocity at any position during the motion.

The maximum velocity occurs when the car's spring passes through the equilibrium position. On substituting the known quantities, the maximum velocity is obtained. The total energy at the equilibrium position is then obtained by substituting the known and calculated quantities.

When the car's spring moves from maximum to half amplitude, the velocity is considered negative. The total energy is calculated by substituting the velocity with other known quantities.

In both cases, the total energy of simple harmonic motion is the same.

## 15.7: Problem Solving: Energy in Simple Harmonic Motion

Simple harmonic motion (SHM) is a type of periodic motion in time and position, in which an object oscillates back and forth around an equilibrium position with a constant amplitude and frequency. In SHM, there is a continuous exchange between the potential and kinetic energy, which results in the oscillation of the object.

Consider the spring in a shock absorber of a car. The spring attached to the wheel executes simple harmonic motion while the car is moving on a bumpy road. The force on the spring is conservative, and the potential energy is stored when the spring is extended or compressed. In this case, the wheel attached to the spring oscillates in one dimension, with the force of the spring acting parallel to the motion. At the equilibrium position, the potential energy stored in the spring is zero. If there are no dissipative forces, the total energy is the sum of the potential energy and the kinetic energy and is expressed as follows:

The total energy in simple harmonic motion remains conserved for the system at every point during the motion and is proportional to the square of the amplitude.

The total energy equation in simple harmonic motion presents a useful relationship between velocity, position, and total mechanical energy. This equation can be used if the problem requires a relation between position, velocity, and acceleration without reference to time. Since the energy conservation equation involves displacement and velocity, one must infer the signs of the displacement and velocity from the situation. For instance, if the body moves from the equilibrium position toward the point of the greatest positive displacement, the displacement and velocity are positive.

Studying the energy in simple harmonic motion is vital for understanding the behavior of oscillating systems in physics and engineering.

• Young, H.D. and Freedman, R.A. (2012). University Physics with Modern Physics . San Francisco, CA: Pearson. pp.446.
• OpenStax. (2019). University Physics Vol. 1 . [Web version]. Pp. 754. Retrieved from https://openstax.org/books/university-physics-volume-1/pages/15-2-energy-in-simple-harmonic-motion

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## Browse Course Material

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• Prof. Yen-Jie Lee

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• Atomic, Molecular, Optical Physics
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## Learning Resource Types

Physics iii: vibrations and waves, lecture 1: periodic oscillations, harmonic oscillators.

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## Lecture Topics

Lecture video: periodic oscillations, harmonic oscillators.

In this lecture, Prof. Lee discusses the mathematical description of the periodic oscillation and simple harmonic oscillators. The first 5 minutes are devoted to course information.

## Lecture Notes

Typed Notes for Lecture 1 (PDF)

Handwritten Notes for Lecture 1 (PDF - 2.2MB)

Chapter 1: Harmonic Oscillation (PDF - 1.4MB)

## Problem Set

Problem Set 1 (PDF)

## Problem Solving Help Video*

Simple Harmonic Motion and Introduction to Problem Solving

## In-class Demonstrations

* Note : This Problem Solving Help video was originally produced as part of a physics course that is no longer available on OCW.

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#### IMAGES

1. Problem Solving on Kinematics of Simple Harmonic Motion

2. Simple Harmonic Motion (16 of 16): Equations, Example Problems

3. Simple Harmonic Motion (12 of 16): Example Problems

4. Simple Harmonic Motion Example Problem

5. Problem solving involving Simple Harmonic Motion Models

6. How To Solve Simple Harmonic Motion Problems In Physics

#### VIDEO

1. Problem Solving Simple Harmonic Motion

2. Simple Harmonic Motion

3. Simple Harmonic Motion (SHM)

4. Simple Harmonic motion| Chapter-wise Question Solving| Score 180 in Physics| NEET Physics| Raja sir

5. Simple Harmonic Motion

6. EM 30 |SIMPLE HARMONIC MOTION PROBLEM 3

1. 15.1 Simple Harmonic Motion

In simple harmonic motion, the acceleration of the system, and therefore the net force, is proportional to the displacement and acts in the opposite direction of the displacement. A good example of SHM is an object with mass m attached to a spring on a frictionless surface, as shown in Figure 15.3.

2. Simple Harmonic Motion Problems for High Schools

Simple Harmonic Motion (SHM) Problems. Problem (1): A .50-kg object is attached to a horizontal spring whose spring constant is k=300 N/m and is undergoing a simple harmonic motion. Calculate its. (a) Period, (b) frequency, (c) angular frequency. Solution: Imagine an object is attached to an unstretched spring, displaces the spring from its ...

3. Equation Overview for Simple Harmonic Motion Problems

Simple Harmonic Motion Energy Considerations Since there is no non-conservative force doing work on the mass as it cycles back and forth the Total Mechanical Energy of the mass is conserved:. where KE = ½•m•v 2 is the kinetic energy of the motion, PE = ½•k•x 2 is the potential energy in the spring. Let's look at these values for the 0.5 second intervals of time as in the table above.

4. 15.2: Simple Harmonic Motion

Example 15.2: Determining the Equations of Motion for a Block and a Spring. A 2.00-kg block is placed on a frictionless surface. A spring with a force constant of k = 32.00 N/m is attached to the block, and the opposite end of the spring is attached to the wall. The spring can be compressed or extended.

5. PDF Chapter 23 Simple Harmonic Motion

1. One of the most important examples of periodic motion is simple harmonic motion (SHM), in which some physical quantity varies sinusoidally. Suppose a function of time has the form of a sine wave function, y(t) = Asin(2πt / T ) (23.1.1) where A > 0 is the amplitude (maximum value).

6. 13.7: Sample problems and solutions

Show that this is simple harmonic motion and determine what Ty's maximum speed will be. Figure 13.7.1 13.7. 1: Playground equipment made of platform connected to two vertical springs. Answer. Exercise 13.7.2 13.7. 2. A torsional pendulum consists of a horizontal rod suspended from a vertical wire.

7. Simple Harmonic Motion: General Solution

The most general solution for simple harmonic motion is . The angle is called the phase of the oscillation, and it is telling us where in the oscillation the problem starts (i.e. when ). For this book, we will only concern ourselves with oscillations that start either at maximum displacement (cosine solution) or with the object at the ...

8. Simple Harmonic Motion and Introduction to Problem Solving

Simple Harmonic Motion and Introduction to Problem Solving Harmonic Oscillators with Damping Driven Harmonic Oscillators Coupled Oscillators without Damping ... Simple Harmonic Motion and Introduction to Problem Solving. Simple Harmonic Motion and Introduction to Problem Solving. Viewing videos requires an internet connection

9. Simple harmonic motion

Simple harmonic motion: Finding frequency and period from graphs Get 3 of 4 questions to level up! Simple harmonic motion: Finding speed, velocity, and displacement from graphs Get 3 of 4 questions to level up! Quiz 1. Level up on the above skills and collect up to 160 Mastery points Start quiz.

10. Problems

Problem 2. A massive bird lands on a taut light horizontal wire. You observe that after landing the bird oscillates up and down with approximately simple harmonic motion with period T. You also notice that during the initial oscillations the maximum height the bird reaches above its position at landing is equal to the distance below its landing ...

11. How To Solve Simple Harmonic Motion Problems In Physics

This physics video tutorial provides a basic introduction into how to solve simple harmonic motion problems in physics. It explains how to calculate the fre...

12. 1. Simple Harmonic Motion & Problem Solving Introduction

View the complete OCW resource: http://ocw.mit.edu/resources/res-8-005-vibrations-and-waves-problem-solving-fall-2012/Instructor: Wit BuszaWe discuss the rol...

13. Simple harmonic motion

The most common form of restoring force is the Hooke's Law. Percisely, a system demonstrates simple harmonic motion when an object's accerleration is porportional to its position and is oppositely directed to the displacement from equilibrium position.

14. Simple harmonic motion

Simple harmonic motion (SHM) is an mechanical example of periodic motion, a motion of an object that regularly repeat itself. If a force is directed toward the equilibrium position, that motion is referred as harmonic motion. Generally, this force directed toward the eqilibrium position, the opposite of t he displacement vector, is often called ...

15. Simple Harmonic Motion and Introduction to Problem Solving

About this Video In this first session, after a brief introduction, we discuss the role problem-solving plays in the scientific method. We then focus on problems involving simple harmonic motion—i.e., on harmonic oscillators with one degree of freedom in which damping (frictional or drag) forces can be ignored.

16. Solved Example Problems for Simple Harmonic Motion (SHM)

Simple harmonic oscillation equation is y = A sin(ωt + φ 0) or y =A cos(ωt + φ 0) EXAMPLE 10.7. Show that for a simple harmonic motion, the phase difference between. a. displacement and velocity is π/2 radian or 90°. b. velocity and acceleration is π/2 radian or 90°. c. displacement and acceleration is π radian or 180°. Solution. a.

17. Simple Harmonic Motion Problem Solving (pendulum, mass on a spring

Examples of solving simple harmonic motion problems.0:00 - Intro0:37 - Problem 15:52 - Problem 28:07 - Problem 3Problem 1 - Having landed on a newly discover...

18. The Calculator Pad: Simple Harmonic Motion Problem Sets

The problems target your ability to use of simple harmonic motion equations combined with force relationships to solve problems involving cyclical motion and springs Click a link to open a publicly-available problem set. If you are a Task Tracker student, open the assignment using the link on the Task Tracker assignment board.

19. PDF MITOCW

Simple Harmonic Motion & Problem Solving Introduction The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT

20. Problem Solving: Energy in Simple Harmonic Motion

15.7: Problem Solving: Energy in Simple Harmonic Motion Simple harmonic motion (SHM) is a type of periodic motion in time and position, in which an object oscillates back and forth around an equilibrium position with a constant amplitude and frequency. In SHM, there is a continuous exchange between the potential and kinetic energy, which ...

21. Simple Harmonic Motion

Graphically represent the position, velocity and acceleration of Simple Harmonic Motion. Use the laws of dynamics to determine the natural angular frequency of a system in the limit of very small displacements from equilibrium. Describe the consequences of conservation of mechanical energy for Simple Harmonic Motion (assuming no dissipation).

22. Lecture 1: Periodic Oscillations, Harmonic Oscillators

Problem Solving Help Video* Simple Harmonic Motion and Introduction to Problem Solving. In-class Demonstrations. SEE IT IN THE LECTURE Air Cart Between Springs: Mass on a Spring: Two Pendulums with Different Amplitudes * Note: This Problem Solving Help video was originally produced as part of a physics course that is no longer available on OCW.