OscillationsChapter 14

OscillationsOscillations are when something repeats with the same period. Oscillations may seem like something that doesn't occur all that often, but they come up surprisingly often

mass on a spring

pendulum

waves - sound, light, ocean, earthquakes

often if something is in equilibrium and then displaced a small amount (stable equilibrium)

SpringsA spring is a simple example of something that oscillates.

The force on a spring is Fs = -kx. When you pull the spring past its equilibrium point the force pulls it back.

When the end of the spring gets back to x=0 it now has velocity so it keeps going. Now at x<0 the force pushes the spring towards x=0 and the story repeats over and over again.

TermsDisplacement - the distance the end of the spring is from equilibrium.

Amplitude - the maximum amount of displacement.

Period - the time it takes to complete one full cycle.

Frequency - the number of cycles per time unit.

f =1

TT =

1

f

Simple Harmonic Motion

Simple harmonic motion is a special case of oscillation where the force is proportional to the displacement.

In this case it is rather straightforward to solve for the objects motion. X

F = ma = �kx a =d2x

dt2

differential equation

Solution needs to be something where the second derivative is the negative of what you started with

d2x

dt2+

k

mx = 0

x = A cos (!t+ �)

Simple Harmonic MotionLet’s see how it works

d2x

dt2= �!2A cos (!t+ �)

dx

dt= �!A sin (!t+ �)

x = A cos (!t+ �)

d2x

dt2+

k

mx = 0 �!2A cos (!t+ �) +

k

mA cos (!t+ �) = 0

True if ! =

rk

m

True for all A and φ, in order to set these two we need to use boundary conditions.For example if we know that at t=0 the spring is at maximum displacement of 5cm then

� = 0 A = 0.05m x(t) = 0.05 cos (!t)

A is the amplitude of the oscillation and φ is called the phase angle it tells us the peak displacement occurs

Angular FrequencySince the sin and cosine functions repeat every 2π, we see that ωT=2π. So

ω is often called the angular frequency to differentiate it from the normal frequency f. The angular frequency is how many radians happen per time instead of how many cycles per time.

Note that the angular frequency is the same as the angular velocity that we introduced with rotational motion.

! =2⇡

T= 2⇡f

SHM

x = A cos (!t+ �)

v =dx

dt= �!A sin (!t+ �)

a =d2x

dt2= �!2A cos (!t+ �)

xmax = A

vmax = !A =

rk

mA

amax = !2A =k

mA

For the simple harmonic oscillator we can express the maximum velocity and acceleration of our oscillator. For φ=0 the maximum velocity occurs when the oscillator has a

displacement of zero, while the maximum acceleration occurs at maximum displacement.

Example 14-4The cone of a loudspeaker oscillates in SHM at a frequency of 262Hz (“middle C”). The amplitude at the center of the cone is A=1.5 × 10-4 m and at t=0, x=A. a) What equation describes the motion of the center of the cone? b) What are the velocity and acceleration as a function of time? c) What is the position of the cone at t=1.00ms (10-3 s)?knownf = 262/s

A = 1.5 × 10-4 m

unknownx(t) = ?

x(0.001) = ?

SHMphysics x(t) = A cos(ωt+φ)since x=A at t=0, φ=0

x(t) = A cos (!t)

x(t) = (1.5⇥ 10�4m) cos (1650t)

! = 2⇡f = (2⇡)(262s�1) = 1650 rad/s

v(t) = ?a(t) = ?

vmax = !A = (1650rad/s)(1.5⇥ 10�4m) = 0.25m/s

v(t) = �vmax sin (!t) = �(0.25m/s) sin (1650t)

amax = !2A = (1650rad/s)2(1.5⇥ 10�4m) = 410m/s2

a(t) = �amax sin (!t) = �(410m/s2) cos (1650t)

x(0.001) = �(1.5⇥ 10�4m) cos ((1650rad/s)(0.001s)) = �1.2⇥ 10�5 m

Energy in a SHO

One of the easiest ways to study a simple harmonic oscillator is to consider energy conservation.

The potential energy of a spring is U(x) = 1/2kx2, combining this with the kinetic energy gives:

E =1

2mv2 +

1

2kx2

Example 14-7A spring stretches 0.150m when a 0.300kg mass is gently attached to it. The spring is then set horizontally and compressed 0.100m from the equilibrium point before being released. Determine a) the total energy, b) the kinetic and potential energies as a function of time.

knownm = 0.300kgA = 0.100m

unknownE = ?

KE = ? U = ?

physicsE =

1

2mv2 +

1

2kx2

Need k, spring was balanced when stretched 0.150m, so F = kx = mg

=> k =mg

x =(0.300kg)(9.8m/s2)

0.150m= 19.6N/m

E =1

2kA2 =

1

2(19.6N/m)(0.100m)2 = 9.80⇥ 10�2J

Example 14-7A spring stretches 0.150m when a 0.300kg mass is gently attached to it. The spring is then set horizontally and compressed 0.100m from the equilibrium point before being released. Determine a) the total energy, b) the kinetic and potential energies as a function of time.

x(t) = A cos (!t) ! =

rk

m=

s19.6N/m

0.300kg= 8.08 rad/s

x(t) = 0.100m cos (8.08t)

v(t) = �!A sin (!t) vmax = !A = (8.08rad/s)(0.100m) = 0.808m/s

v(t) = �(0.808m/s) sin (8.08t)

U(t) =1

2kx2

=1

2(19.6N/m)(0.100m)2 cos2 (8.08t) = (9.8⇥ 10�2J) cos2 (8.08t)

= (9.8⇥ 10�2J) sin2 (8.08t)KE(t) =1

2mv2 =

1

2(0.300kg)(0.808m/s)2 sin2 (8.08t)

Simple PendulumThe simple pendulum is another example of an object that undergoes simple harmonic motion. The simple part just means that we can ignore the mass of the string in the pendulum.

For a simple pendulum the force is now gravity instead of the spring force; however, most everything else is just like it was for the oscillating spring.

mg

θ

mg cos(θ)

L

x

F = �mg sin(✓) = �mgx

L

For small angles the acceleration is in the x direction and we have simple harmonic motion.

Only difference is that now,! =

rg

L

mg sin(θ)

ResonanceWe’ve seen that a spring and a pendulum have frequencies set by their physical attributes.

This is not uncommon in nature, many objects have a natural frequency ω0 that they prefer to oscillate at.

If a force that varies near ω0 is applied to this type of object then even very small forces can keep increasing the amplitude of the oscillation till you get a big effect.

This is called a resonance and it is the same thing that happens when you push a swing. Resonances are a very important physical phenomena which we won’t have time to discuss in more detail.

Homework

Chapter 15 - 25, 28, 30, 45