CHAPTER 10CHAPTER 10

Elasticity and Oscillations Elasticity and Oscillations

Oscillation or vibration = back-and-forth motion.

Repeats itself, hence it is periodic. Vibrations common:

Vocal cords when singing/speakingString/rubber band, Spring Simple Pendulum – Grandfather clock

OscillationsOscillations

An object in stable equilibrium:

• If displaced slightly from equilibrium point, will oscillate about the equilibrium point.

• It will experience a restoring force.

• The restoring force - directly proportional to displacement.

• The restoring force is opposite to displacement. F = -kx

Types of Equilibrium

Simple Harmonic Motion (SHM)Periodic motion in which Magnitude of the restoring force is proportional to displacement.

Restoring force is opposite to the displacement.

Spring (Elastic) force: F = -kx So spring motion is SHM.

Simple Harmonic MotionSimple Harmonic Motion

F(x)

x

SpringsSpringsHooke’s Law:Hooke’s Law: The force exerted by a

spring is proportional to the distance by which the spring is stretched or compressed from its relaxed position.

FX = -k xwhere x is the displacement from the relaxed position and k is the constant of proportionality.

relaxed position

(Equilibrium Position)

FX = 0

xx=0

SpringsSpringsrelaxed

position

FX > 0

xx < 0

x = 0

FX < 0

xx > 0

relaxed position

x = 0

Compressed

Stretched

FX = -kx

x=A F=-kA v=0 a=-kA/m

x=0 F=0 v=-vo a=0

x=0 F=0 v=+vo a=0

x=-A F=kA v=0 a=kA/m

x=A F=-kA v=0 a=-kA/m

+A

t-A

x

• AMPLITUDE (x = A) = Maximum displacement

from equilibrium. • PERIOD (T) = Time to make one complete cycle.

• FREQUENCY (f) = # of cycles per second.

x=A F=-kA v=0 a=-kA/m

U=½kA2 K=0

x=0 F=0 v=-vo a=0

U=0 K=½mvo2

x=0 F=0 v=+vo a=0

U=0 K=½mvo2

x=-A F=kA v=0 a=kA/mU =½kA2 K=0

Total energy of the spring-mass system:

E = U(spring) + K(mass) = ½kx2 + ½mv2

Total Energy

•At equilibrium point, E1 = U + K = ½mvo2 + 0

•At maximum displacement points,

E2 = U + K = ½kA2 + 0

•At any other position where displacement is x and the velocity is vx;

E3 = ½kx2 + ½mvx2

•Conservation of energy: E1 = E2 = E3

½kA2 = ½mvo2 = ½kx2 + ½mvx

2

Total Energy

E1 = E2 = E3

½kA2 = ½mvo2 = ½kx2 + ½mvx

2

½mvo2 = ½kA2

OR vo = (k/m) A

Energy in SHMEnergy in SHM A mass is attached to a spring and set to

motion. The maximum displacement is x = AEnergy = U + K = constant!

= ½ k x2 + ½ m v2

At maximum displacement x = A, v = 0Total Energy = ½ k A2 + 0

At zero displacement x = 0

Total Energy = 0 + ½ mvm2

Conservation of energy: ½ k A2 = ½ m vm2

Hence vo = (k/m) AAnalogy w/ marble in bowl

Energy in SHMEnergy in SHM

m

xx=0

0 x

US

½ k A2 = ½ m vm2

vo = (k/m) A

Sinusoidal Nature of SHM

Sinusoidal equation = equation involving sine

and cosine.x

t

t

v

t

a

A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. At what points during its oscillation is the magnitude of the acceleration of the block biggest?

1. When x = +A or -A

2. When x = 0 (i.e. zero displacement)

3. The acceleration of the mass is constant.

Example #1

A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. At what points during its oscillation is the magnitude of the potential energy (U) of the spring biggest?

1. When x = +A or -A

2. When x = 0 (i.e. zero displacement)

3. Potential energy of the spring is constant.

Example #2

A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. If the mass takes 2 seconds to move from x = 0 to x = A, what is the period of the oscillation?

(A) 0.5s (B) 1s (C) 2s (D) 4s (E) 8s

Example #3

x = 0 x = A

Example #4Example #4A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the total energy (K+U) of the mass and spring a maximum? (Ignore gravity).

1. When x = +A or -A (i.e. max displacement)

2. When x = 0 (i.e. zero displacement)

3. The energy of the system is constant.

+A

t-A

x

Example #5Example #5A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the speed of the block biggest?

1. When x = +A or -A (i.e. max displacement)

2. When x = 0 (i.e. zero displacement)

3. The speed of the mass is constant

+A

t-A

x

Example #6Example #6A spring oscillates back and forth on a frictionless

horizontal surface. A camera takes pictures of the position every 1/10th of a second. Which plot best shows the positions of the mass.

EndPoint EndPointEquilibrium

EndPoint EndPointEquilibrium

EndPoint EndPointEquilibrium

(A)

(B)

(C)

What does moving in a circle have to do with moving back & forth in a straight line ??

m

x = Ax = 0

A

x

y

x = - A

vo

m

• Mass m moving in uniform circular motion, with constant speed vo [Angular velocity = vo/A].

• x-component of position of the mass is same as SHM of the spring-mass system.

m

x = Ax = 0

A

x

y

x = - A

vo

m• Circular motion can be used to describe SHM of the spring-mass system.

• Just pay attention to the x-component of the circular motion.

x =Acos = Acos t

m

x = Ax = 0

A

xy

x = - A

vo

m

•Also, vo = 2A/T = (k/m) A

OR T = 2 (m/k)

• Total energy:

E = = ½mvo2 = ½kA2

OR vo = (k/m) A

vo = A = 2A/T = 2Af

vo = A = (k/m) A

OR = (k/m)

When amplitude (A) is doubled, maximum velocity (vo) will ……

In the final analysis, period (T) depends only on m and k.

When amplitude (A) is doubled, period (T) will ……

+A

t-A

x

¼T ½T ¾T T

Displacement x = Acos t = Acos (2/T)t

Velocity v = -[A]sint

Acceleration a = -[A2]cost

vy = v0cos

vx = -v0sin

m

x = Ax = 0

x

y

x = - A

vo

vo = A

vx = -v0sin = (-A)sin = (-A)sint

A

t-A

v

¼T ½T ¾T T0

x=A F=-kA v=0 a=-kA/m

x=0 F=0 v=-vo a=0

x=0 F=0 v=+vo a=0

x=-A F=kA v=0 a=kA/m

x=A F=-kA v=0 a=-kA/m

+A

t

x

A2

t

Displacement x = [A]cos t

Velocity v = -[A]sint

Acceleration a = -[A2]costa

t

vA

Simple Harmonic Motion:Simple Harmonic Motion:

x(t) = [A]cos(t)

v(t) = -[A]sin(t)

a(t) = -[A2]cos(t)

xmax = A

vmax = vo = A

amax = A2

P10.25/28

The period of oscillation of an object in an ideal spring-and-mass system is 0.50 s and the amplitude is 5.0 cm. What is the speed at the equilibrium point?

T = 0.5 s, A = 5.0 cm, vo = ?

vo = A, = 2/T, vo = (2/T)A

P10.42/49

A body is suspended vertically from an ideal spring of spring constant 2.5 N/m. The spring is initially in its relaxed position. The body is then released and oscillates about its equilibrium position. The motion is described by

y = (4.0 cm)sin[(0.70 rad/s)t]

What is the maximum kinetic energy of the body?

Kmax = ½ mvo2 y =Asin t and = (k/m)

A = 4.0 cm, = 0.7 rad/s, m = k/2

vo = A, = 2/T, vo = (2/T)A

Simple Harmonic Oscillator = 2 f = 2 / Tx(t) = [A] cos(t)v(t) = -[A] sin(t)a(t) = -[A] cos(t)

For a Spring F = kx amax = (k/m) A A(k/m) A

= sqrt(k/m)k

m2 T

m

k

Vertical Mass and SpringVertical Mass and Spring If we include gravity, there are two forces

acting on mass. With mass, new equilibrium position has spring stretched dF = kd – mg = 0

d = mg/k Let this point be y=0F = k(d-y) – mg

= -k ySame as horizontal! SHONew equilibrium position y = d

The Simple PendulumThe Simple Pendulum Simple pendulum – small object (pendulum bob)

attached to the end of a light inextensible string of length L and swung through small displacements.

Motion of the bob back and forth along arc length (s) is a simple harmonic motion.

Restoring force bringing bob back to eqlbm position

F = -mgsin

x

FT

mg

m

s

mgsin

m

Physics 101: Lecture 19, Pg 37

Types of Equilibrium

The Simple PendulumThe Simple Pendulum

Restoring force bringing bob back to eqlbm position = F = -mgsin

For small angles, sin = s/L x/LF = -mg = -mgx/L, i.e. F xCompare with F = -kx, means k mg/LOR k/m = g/L . But k/m = 2

So 2 = g/LSince T = 2(m/k) = 2(L/g)Period of simple pendulum T = 2(L/g) x

FT

mg

m

s

mgsin

L

The Simple PendulumThe Simple Pendulum

Period of simple pendulum T = 2(L/g)

What should the length of a pendulum clock be for it to measure time accurately?

You need period T = 1s = 2(L/g). [L ~ 25 cm]

What will happen

(a)If L is increased? [ie if L > 25 cm]

(b) If L is decreased? [ie if L < 25 cm]

(c) If g is decreased? (eg clock taken to the moon).

A grandfather clock is running too fast.

To fix it, should the pendulum be lengthened or shortened?

  (A) lengthened(B) shortened(C) need to know the period

#1

T = 2(L/g)

A simple pendulum has a period of oscillation of one second here on Earth. On Mars (where g < 9.8 m/s2), the pendulum will have a period of oscillation

  (A) greater than one second.(B) equal to one second.(C) less than one second.

#2

T = 2(L/g)

Two simple pendulums A and B

Same lengths, Mass of A is twice the mass of B.

Equal vibrational amplitudes.

Periods TA and TB

Energies EA and EB

Choose the correct statement:

A)TA = TB and EA > EB

B)TA > TB and EA > EB

C)TA > TB and EA < EB

D)TA = TB and EA < EB

A mass attached to a spring oscillates with a simple harmonic motion. If it takes 28 seconds to make 40 cycles, what is the frequency of the oscillation?

A. B. C. D. E.

0% 0% 0%0%0%

A. 7.0 Hz

B. 1,120 Hz

C. 0.70 Hz

D. 2.80 Hz

E. 1.43 Hz

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A mass on a spring oscillates with SHM. It takes 0.82 sec to move from x = 0 to x =A. What is the period of the oscillation?

A. B. C. D. E.

20% 20% 20%20%20%

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A. 1.22 s

B. 0.21 s

C. 0.41 s

D. 1.64 s

E. 3.28 s

A mass on a spring oscillates with simple harmonic motion. If the amplitude (A) is increased by a factor 3, the maximum kinetic energy of the mass will increase by a factor

A. B. C. D. E. F. G.

0% 0% 0% 0%0%0%0%

A. 1

B. 1/3

C. 1/6

D. 1/9

E. 3

F. 6

G. 9

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A spring-mass system is in simple harmonic motion and the velocity of the mass is plotted against time as shown below. When is the potential energy of the spring at its minimum?

A. B. C. D. E.

48%50%

3%0%0%

A. 2s, 6s

B. 0s, 4s, 8s

C. 0s, 2s, 4s, 6s, 8s

D. 1s, 2s, 3s

E. The potential energy is constant throughout

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