vibrations and waves

Vibrations and Waves

Physics 2053Lecture Notes

m



13.01 Hooke’s Law 13.02 Elastic Potential Energy13.03 Comparing SHM with Uniform Circular Motion 13.04 Position, Velocity and Acceleration as a Function of Time

Topics

13.05 Motion of a Pendulum

Vibrations and Waves (3 of 33)

Hooke’s Law

If an object vibrates or oscillates back and forth over the same path, each cycle taking the same amount of time, the motion is called periodic (T).

m

We assume that the surface is frictionless. There is a point where the spring is neither stretched nor compressed; this is the equilibrium position. We measure displacement from that point (x = 0 ).

x = 0


m

x = 0

Hooke’s Law

The minus sign on the force indicates that it is a restoring force – it is directed to restore the mass to its equilibrium position.

The restoring force exerted by the spring depends on the displacement:

m

x

F

[13.1] kxFs


m

x

F kxF

(a) (k) is the spring constant

(b) Displacement (x) is measured from the equilibrium point

(c) Amplitude (A) is the maximum displacement

(e) Period (T) is the time required to complete one cycle(f) Frequency (f) is the number of cycles completed per second

(d) A cycle is a full to-and-fro motion

Hooke’s Law


If the spring is hung vertically, the only change is in the equilibrium position, which is at the point where the spring force equals the gravitational force.

m

xo okxF

mgEquilibrium

Position

Hooke’s Law


Any vibrating system where the restoring force is proportional to the negative of the displacement

moves with simple harmonic motion (SHM), and is often called a simple harmonic oscillator.

kxF

Hooke’s Law


Potential energy of a spring is given by:

The total mechanical energy is then:

The total mechanical energy will be conserved

2kx

2mvE

22total

Elastic Potential Energy

[13.3] 2

kxPE2

s


If the mass is at the limits of its motion, the energy is all potential.

m

A 2kAPE

2

m

x = 0

vmax

If the mass is at the equilibrium point, the energy is all kinetic.

2mv

KE2max



This can be solved for the velocity as a function of position:

The total energy is, therefore

And we can write: 2

kA2

mvE22

maxtotal

2kx

2mv

2kA 222

where mkAA

mkv 2

max


[13.6] xA mkv 22


The acceleration can be calculated as function of displacement

m

x

F

kxF

kxma

xmka

Amkamax



If we look at the projection onto the x axis of an object moving in a circle of radius A at a constant speed vmax, we find that the x component of its velocity varies as:

This is identical to SHM.

A

vmax

x

θsinvv max

22 xAmkv

22 xA

mkAvmax A

xAθsin22

AxA

mkAv

22

v

Comparing Simple Harmonic Motion with Circular Motion


2

2max

Ax1vv

Therefore, we can use the period and frequency of a particle moving in a circle to find the period and frequency of SHM:

mkAvmax

mkAvmax

TAπ2

kmπ2T

mk

π21

T1f

Comparing Simple Harmonic Motion with Circular Motion

fA2


A mass m at the end of a spring vibrates with a frequency of 0.88 Hz. When an additional 680 g mass is added to m, the frequency is 0.60 Hz. What is the value of m?

Comparing SHM with Uniform Circular Motion (Problem)


A mass on a spring undergoes SHM. When the mass passes through the equilibrium position, its instantaneous velocity

A) is maximum.

B) is less than maximum, but not zero.

C) is zero.

D) cannot be determined from the information given.



A mass is attached to a vertical spring and bobs up and down between points A and B. Where is the mass located when its kinetic energy is a minimum?

A) at either A or B

B) midway between A and B

C) one-fourth of the way between A and B

D) none of the above



A 0.60 kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.13 m. Determine the velocity when it passes the equilibrium point,



A 0.60 kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.13 m. Determine the velocity when it is 0.10 m from equilibrium,

Comparing SHM with Uniform Circular Motion (Problem) con’t

m/s 45.2vmax


A 0.60 kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.13 m. Determine the total energy of the system,


m/s 45.2vmax


A mass is attached to a vertical spring and bobs up and down between points A and B. Where is the mass located when its potential energy is a minimum?

A) at either A or B

B) midway between A and B

C) one-fourth of the way between A and B

D) none of the above



Doubling only the amplitude of a vibrating mass-and-spring system produces what effect on the system's mechanical energy?

A) increases the energy by a factor of two

B) increases the energy by a factor of three

C) increases the energy by a factor of four

D) produces no change



A mass of 2.62 kg stretches a vertical spring 0.315 m. If the spring is stretched an additional 0.130 m and released, how long does it take to reach the (new) equilibrium position again?



Doubling only the spring constant of a vibrating mass-and-spring system produces what effect on the system's mechanical energy?

A) increases the energy by a factor of three

B) increases he energy by a factor of four

C) produces no change

D) increases the energy by a factor of two



A simple pendulum consists of a mass at the end of a lightweight cord. We assume that the cord does not stretch, and that its mass is negligible.

The Simple Pendulum


mg

F

s

x

sinmgF

Lxsin

x L

mgF

Small angles x s

s L

mgF

k forSHM

km2T

Lmgm2 (13.15)

gL2T

L

m

The Simple Pendulum


A simple pendulum consists of a mass M attached to a weightless string of length L. For this system, when undergoing small oscillations

A) the frequency is proportional to the amplitude.

B) the period is proportional to the amplitude.

C) the frequency is independent of the length L.

D) the frequency is independent of the mass M.



The length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12.0° to the vertical. With what frequency does it vibrate? Assume SHM.



The length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12.0° to the vertical. What is the pendulum bob’s speed when it passes through the lowest point of the swing?



The length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12.0° to the vertical. Assume SHM. What is the total energy stored in this oscillation, assuming no losses?



Summary of Chapter 11

For SHM, the restoring force is proportional to the displacement. kxF

The period is the time required for one cycle, and the frequency is the number of cycles per second.

kmπ2T Period for a mass on a spring:

2kx

2mvE

22total

During SHM, the total energy is continually changing from kinetic to potential and back.


A simple pendulum approximates SHM if its amplitude is not large. Its period in that case is:

gLπ2T

The kinematics of a mass/spring system:

mkAA

mkv 2

max

Amkamax

22 xAmkv Velocity

xmka

Acceleration

Summary of Chapter 11


vibrations and waves

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