MSU Physics 231 Fall 2015 1 Physics 231 Topic 7: Oscillations Alex Brown October 14-19 2015

MSU Physics 231 Fall 2015 1

Physics 231Topic 7: Oscillations

Alex BrownOctober 14-19 2015


Key Concepts: Springs and Oscillations

Springs

Periodic MotionFrequency & Period

Simple Harmonic Motion (SHM)

Amplitude & Period

Angular Frequency

SHM Concepts

Energy & SHM

Uniform Circular Motion

Simple Pendulum

Damped & Driven Oscillations Covers chapter 7 in Rex & Wolfson


Springs

xo=0


Hooke’s Law

Fs = -kx Hooke’s law

k = spring constant in unitsof N/m.

x = displacement fromthe equilibrium point xo=0

xo=0


How to find the spring constant k

Fspring = -kx

Fgravity = mg

x = 0

x = d

When the object hanging from the spring is not moving:

∑F = ma = 0

Fspring + Fgravity = 0

-k(d) + mg = 0

k = mg/d

x


Springs

Hooke’s Law:Fs = -kx

k: spring constant (N/m)x: distance the spring is stretched from equilibrium

W = area under F vs x plot = ½kx2

The energy stored in a spring depends on how far the spring has been stretched: elastic potential energy

PEspring = ½ k x2


PINBALL!The ball-launcher spring has aconstant k = 120 N/m.A player pulls the handle 0.05 m.The mass of the ball is 0.1 kg.What is the launching speed?

(PEgravity+PEspring+KEball)pull = (PEgravity+PEspring+KEball)launch

PEspring = ½ k x2 PEgravity = mgh KE = ½ mv2 ME = PE + KE = Constant

mghpull + ½kxpull2 + ½mvpull

2 = mghlaunch + ½kxlaunch2 + ½mvlaunch

2

½120(0.05)2 = 0.1 (9.81) ( 0.05*sin(10o) ) + ½(0.1)vlaunch2

0.15 = 8.5x10-3 + 0.05v2

v =1.7 m/s


Collisions in one dimension given

given m1 and m2 or c = m2 /m1

v1i and v2i

Elastic (KE conserved)

v1f = [(1-c) v1i + 2c v2i] / (1+c)

v2f = [(c-1) v2i + 2v1i] / (1+c)

Inelastic (stick together, KE lost)

vf = [v1i + c v2i] / (1+c)

Momentum p = mvImpulse p = pf - pi = FtConservation of momentum (closed system) p1i + p2i = p1f + p2f


Collisions in one dimension given

given m1 = m2 or c = 1v1i and v2i

Elastic (KE conserved)

v1f = v2i

v2f = v1i

Inelastic (stick together, KE lost)

vf = [v1i + v2i] / 2


Carts on a spring trackk = 50 N/mv = 5.0 m/sm1 = m2 = 0.25 kg

What is the maximum compression of the springif the carts collide a) elastically and b) perfectly inelastic?

A) Conservation of momentum and KE with c = 1 v1f = v v2f = 0 :: v1f = 0 v2f = v = 5

m/sConservation of energy: ½mv2 = ½kx2 (0.5) (0.25) (5)2 = (0.5)(50)x2 x = 0.35

m B) Conservation of momentum only with c = 1 vf = (v1f + v2f)/2 = (v + 0)/2= 2.5 m/sConservation of energy: ½mvf

2 = ½kx2 (0.5)(0.5)(2.5)2 = (0.5)(50)x2 x=0.25 m

m1 m2

v|||||||||||

spring


Hooke’s Law

Fs = -kx Hooke’s law

If there is no friction, the mass continues to oscillate

back and forth.

If a force is proportional to the displacement x, but opposite in direction, the resulting motion of the object iscalled: simple harmonic oscillation


Simple harmonic motion

time (s)

displacement x

c)

b)

a)

A

Amplitude (A): maximum distance fromequilibrium (unit: m)

Period (T): Time to complete one fulloscillation (unit: s)

Frequency (f): Number of completedoscillations per second. f=1/T(unit: 1/s = 1 Herz [Hz])

Angular frequency = 2πf = 2π/T(a)

(b)

(c)


Concept Quiz

time (s)

displacement x

5 m

-5 m

2 4 6 8 10

a) what is the amplitude of the harmonic oscillation?b) what is the period of the harmonic oscillation?c) what is the frequency of the harmonic oscillation?

a) Amplitude: 5 mb) period: time to complete one full oscillation: T = 4sc) frequency: number of oscillations per second f = 1/T = 0.25/s = 0.25 Hz


What the equation?

time (s)

displacement x

5 m

-5 m

2 4 6 8 10

x = A cos(t)

A = 5 T = 2π

= 2π/T = 2π/4 = 1.57 rad/s


Clicker Quiz!A mass on a spring in SHM has amplitude A and period T. What is the total distance traveled after a time interval of 2T?

A) 0B) A/2C) AD) 4AE) 8A

In one cycle covering period T, the mass moves from +A to -A and back: 2A+2A = 4AIn two periods, the mass moves twice this distance: 8A


Energy and VelocityEkin(½mv2) Epot,spring(½kx2) Sum

0 ½kA2 ½kA2

½mv2max

0 ½mv2max

0 ½k(-A)2 ½kA2

conservation of ME: ½mv2max

= ½kA2 so vmax = ±A(k/m)

x=0

x=-A

x=+A


Velocity and acceleration in general

Total ME at any displacement x: ½mv2 + ½kx2

Total ME at max. displacement A: ½kA2

Conservation of ME: ½kA2 = ½mv2 + ½kx2

So: v = ±[(A2-x2)k/m] also F = ma = -kx so a = -kx/m

position x

velocity v Acceleration a

+A 0 -kA/m

0 ±A(k/m) 0

-A 0 kA/m


An ExampleThe maximum velocity and acceleration of mass connected to a spring are vmax=0.95 m/s and amax=1.56 m/s2, respectively. Find the oscillation amplitude.


An ExampleThe maximum velocity and acceleration of mass connected to a spring are vmax=0.95 m/s and amax=1.56 m/s2, respectively. Find the oscillation amplitude. We do not know the mass of the object or the spring constant. So we must find a way to eliminate these variables!

so vmax = ±A(k/m)

so (vmax )2 = A2 (k/m)

so amax = A (k/m)

A = vmax2/amax = (0.95)2/(1.56) = 0.58 m


Harmonic oscillations vs circular motion

r=A

The projection of the position of the circulating object on the x-axis as a function of time is the same as the position of the oscillating spring.

x(t) = A cos


The vx projection of the linear velocity of the rotating object is the velocity of the mass on the spring.


r=A

v

vx

vx(t) = -v sin


But remember for circular motion with constant speed around the circle = t and v = r = Awhere is the “angular frequency”

So x(t) = A cos = A cos(t)

time to complete one circle = T = one period

so T = 2 or = 2/T

x=0

x=-A

x=+A

vx(t) = -v sin = - A sin(t)

r=A


The simple harmonic motion can be described by theprojection of circular motion on the horizontal axis.

x(t) = A cos(t)v(t) = -A sin(t)

• A amplitude of the oscillation• = 2/T = 2f angular frequency• T period • f = 1/T frequency.



time (s)

A

-A

x

A

-A time (s)

v

x = A cos(t)

velocity v = -A sin(t)

m

kv Amax

m

kRemember thus

The angular frequency does not depend on A


acceleration(a)

time (s)

displacement x

A

-A

Newton’s second law: F=ma -kx=ma a=-kx/m

-kA/m

kA/m

Displacement vs Acceleration


time (s)

A

-A

x

time (s)

time (s)

A

-A

x = A cos(t)

v = -A sin(t)

acceleration a = -A 2 cos(t)

v

a

A 2

-A 2


Period of a Spring

Increase m:increase T

Increase k:decrease Tk

mT

22


C

A mass is oscillating horizontally while attached to a springwith spring constant k. Which of the following is true?

a) When the magnitude of the displacement is largest, the magnitude of the acceleration is also largest.b) When the displacement is positive, the acceleration is also positivec) When the displacement is zero, the acceleration is non-zero

Clicker Quiz!


The pendulum

x=+Ax=-A

x=0


The pendulum

Conservation of Energy!At x=+A: PE = mgh KE = ½mv2 = 0At x=0: PE = mg(0) = 0 KE = ½mv2

½mv2 = mghv2 = 2gh velocity at the bottom

y=h

= max = A


The pendulum

Tension at max angle = max T = mg cosmax

Tension at bottom = 0 T = mg + mv2/L

T

mg cosmg sin

L

s


The pendulum

Restoring force: F=-mg sinThe force pushes the mass mback to the central position.T

mg cosmg sin

L

s


The pendulum

Restoring force: F=-mg sinThe force pushes the mass mback to the central position.

sin (radians) if is small

F = - mg also s = L

so: F=-(mg/L)s

T

mg cosmg sin

L

s


pendulum vs spring

L

g

)/(

m

k

sLmgFkxF

The pendulum

Does not depend on massL

g

m

Lmg

m

k

/

spring pendulum


Damped OscillationsNewton’s 1st Law: An object will stay in motion unless acted upon by a force.

SHM is the same: oscillations will last forever in the absence of additional forces.

Common force of friction often “damps” oscillations and brings them to a stop.

In homework A is reduced by some factor f after one period

After one oscillation A1 = f A

So for n oscillations An = fn A

mechanical energy ME = (1/2) k A2


Driven Oscillations

The same can be applied in reverse:

An object NOT in motion can be put into oscillation by applying a force

This is known as driven oscillations.

If the force acts with the same frequency as the as that of the object this is known as resonance.

With driven oscillation, the amplitude grows!


Spring problemA bungee jumper with height h=2 m and mass m = 80 kg leaps from a bridge with height H = 30 m. A bungee cord with spring constant k = 100 N/m attached to his legs. What is the maximum length the cord needs to be if he is to avoid hitting the water below?

Define A = extended “amplitude” of the bungee cordTotal extension = jumper’s height + nominal length of the cord + extended length of the cord = h+L+A Must stop before h+L+A = H = 30m, or A = H-L-h

Use conservation of ME: ½mv2 + ½kx2 + mgHOn top of the bridge: ME = mgHMaximum extension of the cord: ME = ½ kA2 = ½ k(H-L-h)2

ME(bridge) = ME(ground)mgH = ½k(H-L-h) 2 (80 kg)(9.81 m/s2)(30m) = ½(100 N/m)(30m-L-2m) 2

23544 = 50 (28m-L) 2

(28m-L) = sqrt(23544/50) = 21.7 m L = 6.3 m


Spring problemA h=2m tall, m=80 kg bungee jumper leaps from a H=30m bridge with a bungee cord with spring constant k = 100 N/m attached to his legs. What is the frequency of his subsequent SHM?

Frequency f = 1/T= 1/5.62s = 0.18 Hz

= = 5.62 sec


Spring problemA mass of 0.2 kg is attached to a spring with k=100 N/m.The spring is stretched over 0.1 m and released.a) What is the angular frequency () of the corresponding motion?b) What is the period (T) of the harmonic motion?c) What is the frequency (f)?d) What are the functions for x,v and t of the mass as a function of time? Make a sketch of these.

)a =(k/m) = (100/0.2) = 22.4 rad/s)b = 2/T T= 2/ = 0.28 s)c = 2f f=/2 = 3.55 Hz (=1/T)

d) x(t) = Acos(t) = 0.1 cos(22.4t) v(t) = -Asin(t) = -2.24 sin(22.4t) a(t) = -2Acos(t) = -50.2 cos(22.4t)


time (s)

0.1

-0.1

-50.2

50.2

velocity v

a

x

2.24

-2.24

0.28 0.56

0.28

0.28

0.56

0.56


pendulum problemThe machinery in a pendulum clock is keptin motion by the swinging pendulum.Does the clock run faster, at the same speed,or slower if:a) The mass is hung higherb) The mass is replaced by a heavier massc) The clock is brought to the moond) The clock is put in an upward accelerating

elevator?

L m Moon

Elevator

Faster(A)

Same(B)

Slower(C)

L

g


Pendulum QuizA pendulum with length L of 1 meter and a mass of 1 kg, is oscillating harmonically. The period of oscillation is T. If L is increased to 4 m, and the mass replaced by one of 0.5 kg, the period becomes:A) 0.5 T B) 1 TC) 2 TD) 4 TE) 8 T

g

LT 2

Mass does not matter

Tg

L

g

LT 24

42


Spring problemA mass of 1 kg is hung from a spring. The spring stretchesby 0.5 m. Next, the spring is placed horizontally and fixedon one side to the wall. The same mass is attached and thespring stretched by 0.2 m and then released. What isthe acceleration upon release?

1st step: find the spring constant kFspring =-Fgravity or -kd =-mg

k = mg/d = (1)(9.8)/0.5 = 19.6 N/m

2nd step: find the acceleration upon releaseNewton’s second law: F=ma -kx = ma a = -kx/m

a = -(19.6)(0.2)/1 = -3.92 m/s2


Sprng problem

d

x A block with mass of 200 g is placedover an opening. A spring is placed under the opening and compressed by a distance d=5 cm. Its spring constant is 250 N/m. The spring is released and launches the block. How high will it go relative to its rest position (h)? (the spring can be assumed massless)

ME = ½mv2 + ½kd2 + mgx must be conserved.

Initial: v = 0 d = -0.05 m x = 0 m = 0.2 kg

0 + (0.5)(250)(0.05)2 + 0 = 0.3125

Final: v = 0 d = 0 x = h

0 + 0 + (0.2)(9.8)(h) = 1.96 h

Conservation of ME: 0.3125 = 1.96 h so h = 0.16 m

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MSU Physics 231 Fall 2015 1 Physics 231 Topic 7: Oscillations Alex Brown October 14-19 2015