chapter_13

# Ch.13: Oscillations (Mar 31)

## Constants Ch.13

• Standard gravity $g=9.80665\,\frac{\text m}{ {\text s}^2}\approx 9.8\,\frac{\text m}{ {\text s}^2}$

## Equation Sheet Ch.13

Please update if an equation is not included

1. The period $T$ is the time required to complete one cycle of periodic motion
• $T = \frac{1}{f}=\frac{2\pi}{\omega}$ (or, equivalently, $T = \frac{2\pi r}{v}$ in circular motion analogy)
• SI unit: seconds (s)
2. Frequency $f$ is the number of complete oscillations per unit time:
• $f = \frac{1}{T}=\frac{\omega}{2\pi}$
3. Angular frequency $\omega$ (of the point in the shadow analogy):
• $\omega = 2\pi f=\frac{2\pi}{T}$
• SI unit: rad/s = ${\text s}^{-1}$
4. Amplitude (for circular motion)
• $A = r = x_\text{max}$
5. Position vs. time in Simple Harmonic Motion (SHM), also see the table below this list:
• $x = x_\text{max} \cos\big(2\pi\frac{t}{T}+\theta_0\big) = x_\text{max} \cos(\omega t+\theta_0)$
• SI Unit: meters (m)
6. For a rotating turntable (in shadow analogy) angular position of an object:
• $\theta = \omega t+\theta_0$
• SI units: dimensionless or radians
7. The x-coordinate of the radius $r$ connecting the moving point to the center, also see the table below this list:
• $x = r\cos\theta=r\cos\big(\omega t+\theta_0\big)$
8. Velocity in Simple Harmonic Motion (SHM), also see the table below this list:
• $v = x_\text{max}\omega\cos(\omega t+\theta_0)$
• SI unit: $\frac{\,\text m}{\,\text s}$
9. Acceleration in simple harmonic motion, also see the table below this list:
• $a = -x_\text{max}\omega^2 \cos(\omega t+\theta_0)$
• SI unit: $\frac{\,\text m}{\,{\text s}^2}$
10. Angular frequency of a mass on a spring:
• $\omega = \sqrt{\frac{k}{m}}$
11. Period of a mass $m$ attached to a spring of force constant $k$:
• $T = 2\pi\sqrt{\frac{m}{k}}$
• SI Unit: seconds (s)
12. A vertical spring stretches by an amount $y_0$ when a mass is added:
• $y_0 = \frac{mg}{k}$
13. In an oscillating system without friction, energy is conserved:
• $E_1 = E_2$
• $K_1 + U_1 = K_2 + U_2$
14. Energy conservation for a frictionless mass on a spring:
• $\frac{1}{2}mv_1^2 + \frac{1}{2}kx_1^2$ $=\frac{1}{2}mv_2^2 + \frac{1}{2}kx_2^2$ $=\frac{1}{2}mv_\text{max}^2$ $=\frac{1}{2}kx_\text{max}^2$
15. Potential energy of a spring
• $U = \frac{1}{2}kx^2$
16. Maximum value of potential energy of a mass on a spring:
• $U_\text{max}= \frac{1}{2}kx_\text{max}^2=E$, where $E$ is the total energy of the system
17. Kinetic energy as a function of time, also see the table below this list:
• $K =\frac{1}{2}kx_\text{max}^2\sin^2(\omega t+\theta_0)$
18. Potential energy of a simple pendulum (entire mass concentrated at the tip)
• $U = mgL(1 - \cos\beta)\approx \frac{1}{2}mgL\beta^2$
19. Restoring force in a simple pendulum displaced by an arc of length $s$ (corresponding to the angle $\beta$ from the vertical):
• $F= -mg\sin\beta \approx -mg\beta$ $= -mg\frac{s}{L}$
20. Period of a simple pendulum (entire mass concentrated at the tip) with a small amplitude
• $T = 2\pi\sqrt{\frac{L}{g}}$
• SI unit: seconds (s)
21. Period of a physical pendulum (arbitrary mass distribution along the length):
• $T = 2π\sqrt{\left(\frac{L_{\text{c.m.}}}{g}\right)\left(\frac{I}{mL_{\text{c.m.}}^2}\right)}$
• $L_{\text{c.m.}}$ is the distance from the pivot point to the center of mass
• $I$ is the total moment of inertia relative to the pivot point, the sum of individual small parts $\sum m_iL_{\,i}^2$
• Period of a human leg of length $L_{\text{leg}}$:
• assuming the leg is a rod of uniform mass distribution along the length,
• $T = 2\pi\sqrt{\frac{L_{\text{leg}}}{g}}\sqrt{\frac{2}{3}}$
• Note that this result follows from the arbitrary pendulum equation, using
• $L_{\text{c.m.}}=\frac{1}{2}L_{\text{leg}}$
• $I=\frac{1}{3}mL_{\text{leg}}$
22. Maximum values during simple harmonic motion (SHM):
• Maximum velocity: $\;v_\text{max} = x_\text{max}\omega$
• Maximum acceleration: $\;a_\text{max} = x_\text{max}\omega^2$
• Maximum kinetic energy: $\;K_\text{max} =\frac{1}{2}mv_\text{max}^2$ $= \frac{1}{2}m\,x_\text{max}^2\omega^2$ $=U_\text{max}=E$
23. In an underdamped oscillation of initial amplitude $A_0$, mass $m$, damping constant $b$, the amplitude at time $t$ is:
• $A = A_0e^{-\frac{bt}{2m}}$

### Expressions dependent on the starting point

These equations apply only to simple harmonic motion (SHM). The angular frequency $\omega=\frac{2\pi}{T}$.

Starting point: from equilibrium from rest (a) general (b), in terms of sin general (b), in terms of cos
Expression
Initial phase (c) $\theta(0)=-\frac{\pi}{2}$ $\theta(0)=0$ $\theta(0)=\phi-\frac{\pi}{2}$ $\theta(0)=\theta_0$
Coordinate, $x$ $x = x_\text{max}\sin(\omega t)$ $x = x_\text{max}\cos(\omega t)$ $x = x_\text{max}\sin(\omega t+\phi)$ $x = x_\text{max}\cos\!\big(\omega t+\theta_0\!\big)$
Velocity, $v$ $v = x_\text{max}\omega\cos(\omega t)$ $v = -x_\text{max}\omega\sin(\omega t)$ $v = x_\text{max}\omega\cos(\omega t+\phi)$ $v = -x_\text{max}\omega\sin\!\big(\omega t+\theta_0\!\big)$
Acceleration, $a$ $a = -x_\text{max}\omega^2\sin(\omega t)$ $a = -x_\text{max}\omega^2\cos(\omega t)$ $a = -x_\text{max}\omega^2\sin(\omega t+\phi)$ $a = -x_\text{max}\omega^2\cos\!\big(\omega t+\theta_0\!\big)$
Kinetic energy, $K$ $K = \frac{m}{2}x_\text{max}^2\omega^2\cos^2(\omega t)$ $K = \frac{m}{2}x_\text{max}^2\omega^2\sin^2(\omega t)$ $K = \frac{m}{2}x_\text{max}^2\omega^2\cos^2(\omega t+\phi)$ $K = \frac{m}{2}x_\text{max}^2\omega^2\sin^2\!\big(\omega t+\theta_0\!\big)$
Pot. energy, $U$ $U = \frac{m}{2}x_\text{max}^2\omega^2\sin^2(\omega t)$ $U = \frac{m}{2}x_\text{max}^2\omega^2\cos^2(\omega t)$ $U = \frac{m}{2}x_\text{max}^2\omega^2\sin^2(\omega t+\phi)$ $U = \frac{m}{2}x_\text{max}^2\omega^2\cos^2\!\big(\omega t+\theta_0\!\big)$
a) Except when the oscillation is caused by an impact. Use “from equilibrium” column in that case
b) The last two columns are equivalent if one assumes $\phi=\theta_0+\frac{\pi}{2}$, because trigonometrically, $\sin\!\big(\theta+\theta_0+\frac{\pi}{2}\!\big)=\cos\!\big(\theta+\theta_0\!\big)$
c) Initial phase angle measured from the x axis

## Video Ch.13

Video 1.1: Interesting result from different length pendulums. Video 1.2: Engineering application of pendulum. 730 ton ball designed to swing inside of an 101 story building.
Can we calculate the ratio of individual pendulum lengths from this video? How would we do that? A nice write up about the TMD… Interesting quote: “The pinnacle [of the building] may oscillate up to 180,000 times a year due to strong wind loads.” This would be a period $T$ of 175 seconds, or a frequency of 0.006 Hz.

## Math Ch.13

Addition of angles Doubling of angles Squaring of trig functions Small angle $\alpha\ll 1$
$\sin(\alpha+\beta)=$ $\sin\alpha\cos\beta+\cos\alpha\sin\beta$ $\sin(2\alpha)$ $=2\cos\alpha\sin\alpha$ $\sin^2\alpha=$ $\frac{1}{2}\big(1-\cos(2\alpha)\big)$ $\sin\alpha$ $\;\approx\; \alpha$
$\cos(\alpha+\beta)=$ $\cos\alpha\cos\beta-\sin\alpha\sin\beta$ $\cos(2\alpha)$ $=\cos^2\alpha-\sin^2\alpha$ $=2\cos^2\alpha-1$ $\cos^2\alpha=$ $\frac{1}{2}\big(1+\cos(2\alpha)\big)$ $\cos\alpha\;\approx\;$ $1-\frac{\alpha^2}{2}$

## Examples Ch.13

### Problem 1.0

Calculate the following:

1. The frequency of Earth's rotation around its axis
• For the astronomy geeks: Please take into account that the inertial system is referenced to the stars, not the Sun
2. The frequency of Earth's rotation around the Sun
3. The period of your computer processor running at 3.2 GHz

Steps:

1. Rotation around the axis:
1. Period $T = 1\,{\text{day}} \times 24\,\frac{\text{hr}}{\text{day}}\times 60\,\frac{\text{min}}{\text{hr}}\times 60\,\frac{\text s}{\text{min}}$ $= 86,400$ s
2. Frequency $f = \frac{1}{T} =$ $\frac{1}{86,400\,{\text s}} =$ $1.2\times 10^{-5}$ Hz
2. Rotation around the sun:
1. period $T=$ $365\,{\text{days}}\times\frac{24\,{\text{hr}}}{1\,{\text{day}}}\times\frac{3600\,{\text s}}{1\,{\text{hr}}}= 3.156\times10^7$ seconds
2. frequency $f=1/T =$ $1/(3.156\times 10^7\,{\text s}) = 3.2\times 10^{-8}\,$Hz
3. Computer:
1. convert units: 3.2 GHz = $3.2\times 10^9\,$Hz
2. period $T=1/f=$ $1/(3.2\times 10^9\,{\text{Hz}})=$ $3.125\times 10^{-10}\,$seconds = 0.3125 ns
• Question: how far does light travel in this time?
• Student answer (correct!): $3\times 10^8\,{\text{m/s}} \cdot 3.125\times 10^{-10}\,{\text s} = 9.375\times 10^{-2}$ m

### Problem 1.1

A mouse's heart rate increases from 588 beats per minute (BPM) to 612 BPM in two minutes. What is the change in the heart period?

Steps:

1. Convert units to SI:
• $f_1=588\,{\text{BPM}}$ $=\frac{588}{60}\,{\text{Hz}}=9.8\,{\text{Hz}}$
• $f_2=612\,{\text{BPM}}$ $=\frac{612}{60}\,{\text{Hz}}=10.2\,{\text{Hz}}$
2. Convert frequencies to period:
• $T_1=\frac{1}{f_1}=\frac{1}{9.8\,{\text{Hz}}}$ $\approx 0.102\,{\text s}=102\,$ms
• $T_2=\frac{1}{f_1}=\frac{1}{10.2\,{\text{Hz}}}$ $\approx 0.098\,{\text s}=98\,$ms
3. Find the change:
• $\Delta T=T_2-T_1=-4\,$ms
4. Make sense of the answer:
• The mouse heart started to beat faster, its period decreased by 4 ms (by about 4%).

### Problem 1.2

A “pocket balance” (shown to the right) extends by 1/4 of an inch when loaded with a 3 kg mass. Find the frequency with which this mass will oscillate around the equilibrium. Hint: Find the spring constant first.

Steps:

1. Convert units to SI:
• $\Delta l=\frac{1}{4}\,{\text{inch}} = \frac{0.0254\,{\text m}}{4}$ $=0.00635\,{\text m}=6.35\,$mm
2. Find the spring constant from the balance of gravity and spring forces on the mass:
• $mg-k\Delta l=0\;\;\Rightarrow$
• $k=\frac{mg}{\Delta l}=\frac{3\,{\text{kg}}\,\cdot\,9.8\,\frac{\text m}{ {\text s}^2}}{0.00635\,{\text m}}$ $=4630\,\frac{\text N}{\text m}$
3. Find the frequency:
• $f =\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ $\approx\frac{1}{6.283}\sqrt{\frac{4630\,\frac{\text N}{\text m}}{3\,{\text{kg}}}}$ $=6.25\,$Hz
4. Make sense of the answer:
• This is still slow enough to see by eye. Anyone has such a scale at home?

### Problem 1.2A

On a similar device, a baby in Mali is being weighed, held by his grandmother. If she were to let go of the infant, he would bounce up and down 3 times per second. Find out how far down does the infant pull the scale when not bouncing. Can you find the baby's weight from these data?

Steps (assuming k is given in the previous problem):

1. Take k from previous question:
• $k=4630$ N/m.
• Frequency $f = 3\,$Hz
2. Find mass of baby using known k and f.
1. $\;\;f= \frac{1}{2\pi}\sqrt{\frac{k}{m}}$
:
2. $\;\;3\,{\text{Hz}} = 0.159 \sqrt{\frac{4630\;{\text{N/m}}}{m}}$
:
3. $\;\;9\,{\text{Hz}}^2 = 0.025 \frac{4630\;{\text{N/m}}}{m}$
:
4. $\;\;356\,{\text{Hz}}^2 = \frac{4630\;{\text{N/m}}}{m}$
:
5. $\;\;m = 13$ kg - pretty healthy baby!
:
3. use weight equation $\omega = \sqrt{\frac{k}{m}}$
:
1. $\;\;\omega = \sqrt{\frac{4630\;{\text{N/m}}}{13\,{\text{kg}}}}$
:
2. $\;\;\omega = 18.86$ s$^{-1}$
• Step 3 is just not right: the weight (in physics) is defined as the force with which a body is acting on its support.
• In equilibrium, it is always $W=mg$, so should be about $W\approx130$ N
• The equation in Step 3 above is for the angular frequency of the rotating object whose shadow maps onto our oscillation!
Prof. Nicholas Kuzma 2015/04/02 23:45

Challenge yourself: can you outline the steps for the same problem if the spring constant k is not known?

### Problem 1.3

It takes 5 lb of force to charge a small spring-loaded toy gun. If fired straight up, it shoots a 10-g projectile 40 feet high. Find the frequency of oscillation if the projectile gets stuck to the spring.

Steps:

1. Convert all units to SI:
• 10 g = 0.01 kg
• 40 ft = $40\,{\text{ft}}\times 12\,\frac{\text{inch}}{\text{ft}}\times 0.0254\,\frac{\text m}{\text{inch}}$ $=12.19\,$m
• 5 lb of force = $5\,{\text{lb}}\cdot 0.454\,\frac{\text{kg}}{\text{lb}}\cdot 9.8\,\frac{\text m}{ {\text s}^2}$ $=22.2\,$N
2. Use the charging force to relate the spring compression $\Delta l$ to the spring constant $k$:
• $k\cdot|\Delta l|=F=22.2\,$N
3. Instead of getting bogged down in kinematics, use the conservation of energy:
• Pot. Energy of loaded spring (before the shot) =
= Kin. Energy of projectile (right after the shot) =
= Pot. energy of projectile (at max. height)
• $\frac{1}{2}k\,(\Delta l)^2=mgh$
4. We have two equations with two unknowns (k and $\Delta l$). Let's square both sides of the force equation and divide it by the equation above:
• $\frac{k^2(\Delta l)^2}{\frac{1}{2}k\,(\Delta l)^2}=\frac{F^2}{mgh}$ $\;\Rightarrow$
• $2k=\frac{F^2}{mgh}\;\Rightarrow$
• $k=\frac{F^2}{2mgh}$ $=\frac{(22.2\,{\text N})^2}{2\,\cdot\,0.01\,{\text{kg}}\,\cdot\,9.8\,\frac{\text m}{ {\text s}^2}\,\cdot\,12.19\,{\text m}}$ $=206\,\frac{\text N}{\text m}$
5. Using this spring constant and the mass, find the frequency:
• $f =\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ $\approx\frac{1}{6.283}\sqrt{\frac{206\,\frac{\text N}{\text m}}{0.01\,{\text{kg}}}}$ $=22.8\,$Hz
6. Make sense of the answer:
• This is at the lower range of what human ear can detect: “Wrrrrrr!”
• The compression $|\Delta l|$ comes out to be 0.108 m, or about 4 inches. Quite reasonable for a toy gun.

### Problem 1.4

Two pendulums (or, pendula) are made of identical 1 kg masses suspended on two weightless strings, 40.0 and 40.5 cm in length. If these pendulums are deflected from vertical by 5 cm and released at the same time, how long will it take for them to get completely “out of step” with each other?

Steps:

1. Convert relevant quantities to SI units:
• $l_1$ = 40.0 cm = 0.400 m
• $l_2$ = 40.5 cm = 0.405 m
• masses and (small) deflections are irrelevant for finding periods and timing of oscillations
2. Find each period:
• $T_1=2\pi\sqrt{\frac{l_1}{g}}$ $\approx 6.283\sqrt{\frac{0.400\,{\text m}}{9.8\,\frac{\text m}{ {\text s}^2}}}=1.269\,$s
• $T_2=2\pi\sqrt{\frac{l_2}{g}}$ $\approx 6.283\sqrt{\frac{0.405\,{\text m}}{9.8\,\frac{\text m}{ {\text s}^2}}}=1.277\,$s
3. By the time $t$ the two pendulums are “out of step”, one would have completed an extra 1/2 periods compared to the other:
• $t = \left(n+\frac{1}{2}\right)T_1=nT_2$
4. Use this relationship to solve for $n$, then find $t$:
• first, open the parentheses: $t= nT_1+\frac{1}{2}T_1=nT_2$
• then, isolate $n$ by combining terms: $\frac{1}{2}T_1=nT_2-nT_1=n\,(T_2-T_1)$
• lastly, $n=\frac{T_1}{2(T_2-T_1)}$ $=\frac{1.269\,{\text s}}{2\,\cdot\,8\times 10^{-3}\,{\text s}}$ $=79$
• $t=nT_2$ $=79\cdot1.277\,{\text s}\approx 100\,$s
5. Make sense of the answer:
• This is much longer than the periods of oscillations, but short enough to observe the effect before the oscillations die down due to friction

### Problem 1.5

A simple pendulum of length $l=1\,$m is deflected from the vertical by $\beta_\text{max}=5^\circ$ and released from rest. Find the velocity when the pendulum passes $\beta=4^\circ$, 3$^\circ$, 2$^\circ$, 1$^\circ$, and 0$^\circ$ positions.

Steps:

1. Ignoring friction, energy is conserved within the system. Therefore, we can relate the kinetic energy (KE) of the pendulum, to the potential energy:
• $\frac{1}{2}mv^2 + mgh =$constant
• And we know that when $mgh$ is at its maximum, KE = 0.
2. Build equations using these relationships, with the goal of solving for velocity $v$. In this case, we want to find the velocity of the pendulum as it passes from the position $\beta_\text{max}$ to $\beta_x$, or as the pendulum swings from a height of $h_\text{max}$ to $h_x$:
• $\frac{1}{2}mv^2 + mgh_x= mgh_\text{max}$
• $\frac{1}{2}mv^2 = mgh_\text{max} - mgh_x$
• $v^2 = 2g \big(h_\text{max}-h_x\big)$
3. The figure to the right gives us the relationship for $h$ in terms of the pendulum length $l$ and $\cos\beta_x$.
• Substitute for $h_x$:
• $v^2$ $= 2g\left[l(1-\cos\beta_\text{max})-l(1-\cos\beta_x)\right]$ $=2g\left[l-l\cos\beta_\text{max}-l+l\cos\beta_x\right]$ $=2gl\,(\cos\beta_x-\cos\beta_\text{max})$
4. Evaluate $v$ for each position:
• $v_{4^\circ}$ $= \sqrt{2g l\,(\cos 4^\circ-\cos 5^\circ)}$ $= 0.164 \frac{\text m}{\text s}$
• $v_{3^\circ}$ $= \sqrt{2g l\,(\cos 3^\circ-\cos 5^\circ)}$ $= 0.218 \frac{\text m}{\text s}$
• $v_{2^\circ}$ $= \sqrt{2g l\,(\cos 2^\circ-\cos 5^\circ)}$ $= 0.250 \frac{\text m}{\text s}$
• $v_{1^\circ}$ $= \sqrt{2g l\,(\cos 1^\circ-\cos 5^\circ)}$ $= 0.268 \frac{\text m}{\text s}$
• $v_{0^\circ}$ $= \sqrt{2g l\,(\cos 0^\circ-\cos 5^\circ)}$ $= 0.273 \frac{\text m}{\text s}$
5. Remember, that on some calculators punching $\cos 5^\circ$ will actually produce $\cos(5\,\text{rad})$ and give 0.284 instead of 0.996.
• In that case, need to calculate $\cos\left(\frac{\pi}{180^\circ}5^\circ\right)$

## HW Questions Ch.13

Professor, I was reviewing homework problems from chapter 13 and came across one that used the formula, T=2π/square root K/m. But I thought during last lecture you mentioned it could be m/k. Is that true? And if so, in what situations would if differ? — Nichole

• Well, there are two equivalent ways of writing the same formula:
• $T=\frac{2\pi}{\sqrt{\frac{k}{m}}}\Leftrightarrow T=2\pi\sqrt{\frac{m}{k}}$
• What's important (to make sure you are correct), is that when the mass increases, the period should get longer (increase), and when the spring gets stiffer (higher $k$), the motion should become faster, that is, $T$ should decrease.
Prof. Nicholas Kuzma 2015/04/20 21:46

### Problem 54

A 0.55-kg block slides on a frictionless horizontal surface with a speed of 1.1 m/s . The block encounters an unstretched spring and compresses it 21 cm before coming to rest. (Actual numbers may vary)

#### part A

What is the force constant?

• I tried using $\frac{1}{2}mv^2= \frac{1}{2}kx^2$, but I couldn't figure out what x was. Then, I tried $k=\frac{mg}{x}$ and that wasn't the right answer either. — Brittany D.
• You need to be clear what the meaning of equations is. Your first attempt was actually correct, you just need to be able to interpret your variables. You are using the law of conservation of energy, so you are equating the energy at one point in time to the energy at another point. What are these two points in time? The first moment must be when the block “encounters” the spring. What are the kinetic and potential energies there (Hint: the spring is not compressed at that point, $x=0$). The second moment that you are equating is when the block “comes to rest”, if only for a split second, before bouncing back. What are the kinetic and the potential energies at that point? (Hint: “comes to rest” means $v=0$.) If you follow this logical path to the end, you'll realize that your x is the compression of the spring that is given. — Prof. Nicholas Kuzma 2015/04/07 00:06

#### part B

For what length of time is the block in contact with the spring before it comes to rest?

• I looked at the homework solutions posted in D2L and for the solution, I don't understand why it states that “the period corresponds to one-fourth of a period.” Where does the 1/4-th come from? I know that the spring was compressed 0.25 m so are we assuming that the period is originally 1 meter and that's where the 1/4 comes from?
• If you are trying to map the motion described in the problem onto an oscillation, then, at the beggining, when the encounter happens, the potential energy of the spring is zero (unstretched spring), that means you start at an equilibrium point. When the block comes to rest, its velocity (and so its kinetic energy) is zero, that means you end up at the $x_\text{max}$ (maximum distance from equilibrium) position. Since a typical oscillation cycle goes like this: $0 \rightarrow x_\text{max}\!\rightarrow 0 \rightarrow -x_\text{max}\!\rightarrow 0\ldots$ etc, the time between successive zeros (passing the equilibrium point) is half the period, and the time between a zero (equilibrium point) and $x_\text{max}$ is half that, or 1/4 period. If you plot a $\sin(x)$ function, it will be obvious. So, this is how you get 1/4 period.