# Simple Harmonic Motion

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## Contents

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## Formulas

$\omega\ = \frac{2 \pi\ }{T}$

• w = angular velocity in rad per second. It's like we are pretending that the oscillator is rotating. If it goes back and forth and back again in 1 second, then its angular velocity is 2 pi radians per second.
• T = period in seconds. Time to complete one complete cycle.

$x = x_0 \sin\ \! \! \, \omega\ \! \! \, t ; \quad x=x_0 \cos\ \! \! \, \omega\ \! \! \,t$

$v = v_0 \cos\ \! \! \, \omega\ \! \! \, t ; \quad v=-v_0 \sin\ \! \! \, \omega\ \! \! \,t$

• v = velocity of the oscillator at time t in m/s
• vo = maximum velocity in m/s. Occurs at x = 0
• x = position of the oscillator at time t in m
• xo = maximum displacement. AKA amplitude. in m

Whether you use the left or right equations just depends on how you start the oscillator in motion. If at t = 0 it is moving in the positive direction starting at equilibrium (x = 0) then use the left side, and if it starts at positive maximum amplitude with no velocity, then use the right side equations. Otherwise, use your vast knowledge of trig to figure out the setup. Be sure that your calculator is in RADIANS.

$v = \pm\omega\sqrt{(x_0\ ^2 - x^2)}$

This is a formula for getting the speed (velocity without direction) of the oscillator when we know how far it is from equilibrium (x = 0) See above for what all the variables are. Of course for any position, it could be there on the way in or out, so just the position alone cannot tell us the velocity, only the speed. Note that this is useful for finding vo - the maximum speed. vo occurs of course when x = 0, so if you plug in 0 for x you get that vo = wxo

$E_k = \frac{1}{2} m \omega^2 (x_0\ ^2 - x^2)$

This gives you the kinetic energy (Ek) at a particular position x if you know the amplitude xo and the mass m (in kg). If you set this equal to kinetic energy (1/2mv2) you get the equation before it solving for v. Remember, that the oscillator has potential energy (1/2kx2 - not in the data packet) and kinetic energy (1/2mv2) which is in the data packet. The potential is of course the total energy (see below) minus the kinetic energy.

$E_{K (max)} = \frac{1}{2} m \omega^2 x_0\ ^2$

$E_T = \frac{1}{2} m \omega^2 x_0\ ^2$

Basically the same equation twice. The one on the left side is the maximum kinetic energy, which is the same as the oscillator's total energy, (the right hand equation) because maximum kinetic energy occurs at the equilibrium point (x = 0) where there is no potential energy.

## Problems

### 1. a) What is the angular velocity of a Simple Harmonic Oscillator with a period of 2.37 seconds? b) If the oscillator has an amplitude of 0.150 m, what is the maximum velocity it reaches?

$\omega\ = \frac{2 \pi\ }{T}$

$v = \pm\omega\sqrt{(x_0\ ^2 - x^2)}$

Since we're just looking for the maximum velocity, we can get rid of the x2, making the formula shortened to:

vo = ω*xo
vo = (2.65 rad/s)*(0.150 m) vo = 0.3975 = 0.398 m/s

### 2. A SHO has a maximum velocity of 0.870 m/s with an amplitude of 8.50 cm. a) What is its angular velocity? b) What is its period of motion? c) What is its frequency?

$v = \pm\omega\sqrt{(x_0\ ^2 - x^2)}$

We're going to do the same thing that we did in problem 1B, shortening the formula down to vo = ω*xo.

0.870 m/s = ω*(.085 m)

$\omega\ = \frac{2 \pi\ }{T}$

T = .6138744 s = .614 s

Frequency is measure in Hz, which is (1/s). With our given amount of time, .614s, we can find frequency simply by diving 1 by our seconds.

1 / (.614 s) = 1.63 Hz

### 3. A tuning fork vibrates at 256 Hz with an amplitude of 2.10 mm (2.10 x 10-3 m). a) What is its period of motion? b) What is its angular velocity? c) What is its maximum velocity? d) What is its velocity when it is 1.50 mm from equilibrium?

First we convert Hz to seconds, just like problem 2C, only backwards. Hz = 1/seconds.

256 = 1/t
t = .003906 s = 3.91E-3 s

$\omega\ = \frac{2 \pi\ }{T}$

3.91E-3 = 2π/ T

For 3C we will use the vo = ω*xo we derived.

vo = 3.37784 m/s = 3.38 m/s

$v = \pm\omega\sqrt{(x_0\ ^2 - x^2)}$

v = (1610 rad/s)*[(2.1E-3 m)2-(1.5E-3 m)2]1/2
v = 2.36 m/s

### 4. A simple harmonic oscillator has an equation of motion (in m) of x = 2.75sin(3.17t) a) What is its amplitude of motion? b) What is its angular velocity? c) What is its position at t = 14.2 seconds? d) What is its maximum velocity? A different SHO has a period of 2.8 seconds, and an amplitude of 3.4 m. e) Write the equation for its position, and f) write an equation for its velocity vs. time.

$x = x_0 \sin\ \! \! \, \omega\ \! \! \, t$

2.75 m is the amplitude, 3.17 rad/s is the angular velocity.

x = 2.75 * sin(3.17 * 14.2)
x = 2.35998 m = 2.36 m

To find maximum velocity, we will use our derived formula, vo = ωxo.

vo = 8.7175 m/s = 8.72 m/s

In order to write a new equation for position, we need both amplitude and angular velocity. Our amplitude is given as 3.4 m. Our angular velocity can be solved with the equation:

$\omega\ = \frac{2 \pi\ }{T}$

ω = (2π) / (2.8 s)

Combining our ω and xo, we get the equation x = 3.4sin(2.2t)

To find the velocity equation, we now need to find maximum velocity through our angular velocity and amplitude with vo = ωxo.

vo = 7.5 m/s

If position is a sin equation, velocity switches to cosine. This makes our velocity equation v = 7.5cos(2.2t)

### 5. A simple harmonic oscillator has an equation of motion (in m) of x = 0.450sin(45.1t) a) What is its position at t = 156.7 seconds? b) What is its maximum velocity? c) What is its speed when it is 0.34 m from equilibrium? d) What is the equation for its velocity? e) What is its velocity at t = 156.7 seconds?

x = 0.450sin(45.1 * 156.7)
x = -0.44444 = -0.444 m

vo = ωxo

vo = 20.295 m/s = 20.3 m/s

$v = \pm\omega\sqrt{(x_0\ ^2 - x^2)}$

v = (45.1 rad/s)[(0.450 m)2 - (0.34 m)2]1/2
v = 13.2949 m/s = 13.3 m/s

The equation with our maximum velocity and angular velocity will look like v = 20.3cos(45.1t)

v = 20.3cos(45.1*156.7)
v = 3.18055 m/s = 3.18 m/s

### 6. A SHO has a period of 0.0318 seconds, and has a mass of 0.185 kg that moves with an amplitude of 0.0650 m. a) What is the total energy of this system? (find ω first) b) What is the maximum kinetic energy?, c) what is the maximum potential energy? d) When the mass is displaced only 0.0340 m from equilibrium, what is the kinetic energy? e) What is the potential energy when it is that far from equilibrium?

$\omega\ = \frac{2 \pi\ }{T}$

ω = 2π / (0.0318 s)

$E_T = \frac{1}{2} m \omega^2 x_0\ ^2$

ET = (.5)*(0.185 kg)*(197.5844 rad/s)2*(0.0650 m)2
ET = 15.2572 J = 15.3 J

This answer pertains to both a), b) and c), because the equations for maximum kinetic, potential and total energy are the same.

$E_k = \frac{1}{2} m \omega^2 (x_0\ ^2 - x^2)$

Ek = (.5)*(197.5844 rad/s)2*[(0.0650 m)2-(0.450 m)2]
Ek = 11.0827 J = 11.1 J

To find potential, we take kinetic and subtract it from the total energy.

ET = Ep + Ek
15.3 J = Ep + 11.1 J
Ep = 4.2 J

### 7. A SHO has an energy of 348 J, has a mass of 2.37 kg, and moves with an amplitude of 0.670 m. What is its period of motion? What is its maximum speed? What is its speed when it is only 0.450 m from equilibrium? What is its possible equation of position vs. time? What is its possible equation of velocity vs. time?

Again, we have to solve for angular velocity before doing anything else. With our given values, we will have to use the equation:

$E_T = \frac{1}{2} m \omega^2 x_0\ ^2$

348 J = (.5)*(2.37 kg)*(ω)2*(0.670 m)2

We can now solve for period with our angular velocity.

$\omega\ = \frac{2 \pi\ }{T}$

25.57736 rad/s = 2π / (T)
T = 0.245654 s = 0.246 s

vo = ωxo

vo = 17.13683 m/s = 17.1 m/s

$v = \pm\omega\sqrt{(x_0\ ^2 - x^2)}$

v = (25.57736 rad/s)*[(0.670 m)2 - (0.450 m)2]1/2
v = 12.69627 m/s = 12.7 m/s

Using our maximum position and velocity along with our angular velocity, we can find both equations at once.

x = 0.67sin(25.6t)
v = 17.1cos(25.6t)

### 8. A simple harmonic oscillator with a mass of 0.551 kg has an equation of motion of x = 0.160sin(2.31t) i) What is its total energy? ii) What is its maximum velocity? At time = 13.5 seconds, find: a) the position of the oscillator, b) the velocity of the oscillator c) the kinetic energy of the oscillator and d) its potential energy.

$E_T = \frac{1}{2} m \omega^2 x_0\ ^2$

ET = (.5)*(0.551 kg)*(2.31 rad/s)2*(0.160 m)2
ET = 0.0376344 J = 0.0376 J

vo = ωxo

vo = 0.3696 m/s = 0.370 m/s

The original equation, x = 1.60sin(2.31*13.5)
x = -0.036621 m = -0.0366 m

$v = \pm\omega\sqrt{(x_0\ ^2 - x^2)}$

v = (2.31 rad/s)*[(0.160 m)2 - (0.0366 m)2]1/2
v = 0.359800 m/s = 0.360 m/s

$E_k = \frac{1}{2} m \omega^2 (x_0\ ^2 - x^2)$

Ek = (.5)*(0.551 kg)*(2.31 rad/s)2*[(0.160 m)2 - (0.0366 m)2]
Ek = 0.0356652 J = 0.0357 J

ET = Ep + Ek
0.0376 J = Ep + 0.0357 J
Ep = 0.0019 J

### 9. If you set the equation total energy of an SHO given in your data packet equal to the potential energy stored in a spring (not in the data packet) you get this: 1/2mω2xo2 = 1/2kxo2 Solve for ω in terms of k and m. (aha!)

The first thing you should notice is that both the xo2 and the 1/2 cancel each other out, leaving mω2 = k. Solving for terms of ω is simple algebra from that point on.

ω = (k / m)1/2