Oscillations

Unit 7

Lesson 1 : Simple Harmonic Motion

Fs = -kx

(Hooke’s Law)

Fs is a restoring force because it always points toward the

equilibrium position (x = 0)

Applying Newton’s Second Law :

F = max

-kx = max

ax = -k

mx

Simple Harmonic Motion

An object moves with simple harmonic motion whenever its acceleration is proportional to its

position and is oppositely directed to the displacement from equilibrium.

Example 1

A block on the end of a spring is pulled to position x = A and released. In one full cycle of its motion, through what

total distance does it travel ?

_____ A/2

_____ A

_____ 2A

_____ 4A

a =dv

dt=

d2x

dt2Since ,

d2x

dt2= -

k

mx

Notice that the acceleration of the particle in SHM is not constant. It

varies with position x.

If we call k/m = 2,

d2x

dt2= -2x

We need a function x(t) whose second derivative is the same as the original function

with a negative sign and multiplied by 2.

d2x

dt2= -2x

(second-order differential equation)

x(t) = A cos(t + )

One possible solution is :

Proof

dx

dt= A cos(t + )

d

dt= -A sin(t + )

d2x

dt2= -A sin(t + )

d

dt= -2A cos(t + )

d2x

dt2= -2x

Amplitude (A) : maximum value of the position of the particle in either the positive or negative direction.

x(t) = A cos(t + )

Angular Frequency () : number of oscillations per second.

Since k/m = 2,

= km

= 0

Phase Constant () : initial phase angle. This is determined by the position of the particle at t = 0.

If particle is at maximum position x = A at t = 0, the

phase constant = 0.

Pen traces out cosine curve x(t) = A cos(t + )

Period (T) : the time interval required for the particle to go through one full cycle of its motion.

T =2

Frequency (f) : the inverse of period. The number of oscillations that the

particle undergoes per second.

f =1

T

Since T = 2/,

f =2

The Hertz (Hz) is the SI unit for frequency.

Since f = /2,

= 2f =2T

Since = k/m and T = 2/ ,

T = 2 m

k

f =k

m

1

2

depend only on mass and

spring constant

Velocity in Simple Harmonic Motion

dx

dt= A cos(t + )

d

dt= -A sin(t + )

v = -A sin(t + )

Acceleration in Simple Harmonic Motion

d2x

dt2= -A sin(t + )

d

dt= -2A cos(t + )

a = -2A cos(t + )

Maximum Speed in Simple Harmonic Motion

vmax = +/- A

Since sine and cosine oscillate between +/- 1,

vmax =km

A (magnitude only)

Maximum Acc. in Simple Harmonic Motion

amax = +/- 2A

amax =km A (magnitude only)

position vs. time

velocity vs. time

acceleration vs. time

phase difference is /2 rad or 90o

when x at max or min, v = 0

when x = 0, v is max

phase difference is rad or 180o

when x at max, a is max in opposite direction

Example 2

An object oscillates with simple harmonic motion along the x-axis. Its position varies

with time according to the equation

x = (4.00 m) cos(t + /4)

where t is in seconds and the angles in the parentheses are in radians.

a) Determine the amplitude, frequency, and period of the motion.

b) Calculate the velocity and acceleration of the object at any time t.

c) Using the results of part b, determine the position, velocity, and acceleration of the object at t = 1.00 s.

d) Determine the maximum speed and maximum acceleration of the object.

e) Find the displacement of the object between t = 0 and t = 1.00 s.

Example 3A 200 g block connected to a light spring for

which the force constant is 5.00 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 5.00 cm from equilibrium and

released from rest, as shown below.

a) Find the period of its motion.

b) Determine the maximum speed of the block.

c) What is the maximum acceleration of the block ?

d) Express the position, speed, and acceleration as functions of time.

e) The block is released from the same initial position, xi = 5.00 cm, but with an initial velocity of vi = -0.100 m/s. Which parts of the solution change and what are the new answers for those that do change ?

A 2 kg block is dropped from a height of 0.45 m above an uncompressed spring, as shown above. The spring has an elastic constant of 200 N/m and negligible mass. The block strikes the end of the

spring and sticks to it.

Example 4 : AP 1989 # 3

a) Determine the speed of the block at the instant it hits the end of the spring.

b) Determine the period of the simple harmonic motion that ensues.

c) Determine the distance that the spring is compressed at the instant the speed of the block is maximum.

d) Determine the maximum compression of the spring.

e) Determine the amplitude of the simple harmonic motion.

Example 5 : AP 2003 # 2

An ideal spring is hung from the ceiling and a pan of mass M is suspended from the end of the

spring, stretching it a distance D as shown above. A piece of clay, also of mass M, is then

dropped from a height H onto the pan and sticks to it. Express all algebraic answers in terms of

the given quantities and fundamental constants.

a) Determine the speed of the clay at the instant it hits the pan.

b) Determine the speed of the pan just after the clay strikes it.

c) Determine the period of the simple harmonic motion that ensues.

d) Determine the distance the spring is stretched (from its initial unstretched length) at the

moment the speed of the pan is a maximum. Justify your answer.

The clay is now removed from the pan and the pan is returned to equilibrium at the end of the spring. A rubber ball, also of mass M, is dropped from the same height H onto the pan, and after the collision

is caught in midair before hitting anything else.

e) Indicate below whether the period of the resulting simple harmonic motion of the pan is greater than, less than, or the same as it was in part c.

____ Greater than ____ Less than ____ The same as

Justify your answer.

Lesson 2 : Energy in Simple Harmonic Motion

KE of Block

KE = ½ mv2

Since v = -A sin(t + ),

KE = ½ m2A2 sin2(t + )

Elastic PE Energy Stored in Spring

U = ½ kx2

Since x = A cos(t + ),

U = ½ kA2 cos2(t + )

Total Mechanical Energy of Simple Harmonic Oscillator

E = KE + U

E = ½ m2A2 sin2(t + ) + ½ kA2 cos2(t + )

E = ½ kA2 [sin2(t + ) + cos2(t + )]

Since 2 = k/m,

Since sin2 + cos2 = 1,

E = ½ kA2

E = ½ kA2

The total mechanical energy of a simple harmonic oscillator is a constant of the

motion and is proportional to the square of the amplitude.

U is small when KE is large, and vice versa.

KE + U = constant

E = ½ kA2

Since E = KE + U,

½ kA2 = ½ mv2 + ½ kx2

Solving for v,

v = +/-km

(A2 – x2)

Since =

km

,

v = +/- A2 – x2

velocity of simple

harmonic oscillator

Example 1

The amplitude of a system moving in simple harmonic motion is doubled.

Determine the change in the

a) total energy

b) maximum speed

c) maximum acceleration

d) period

Example 2

A 0.500 kg cart connected to a light spring for which the force constant is

20.0 N/m oscillates on a horizontal, frictionless air track.

a) Calculate the total energy of the system and the maximum speed of the cart if

the amplitude of the motion is 3.00 cm.

b) What is the velocity of the cart when the position is 2.00 cm ?

c) Compare the kinetic and potential energies of the system when the position is 2.00 cm ?

Lesson 3 : Comparing Simple Harmonic Motion with Uniform Circular Motion

As the turntable rotates with constant angular speed, the shadow of the ball moves back and forth in simple harmonic motion.

Reference Circle

Simple harmonic motion along a straight line can be represented by the projection of uniform circular motion along a diameter of a reference circle.

t = 0

t > 0

Q is the projection

of P

x coordinate is x(t) = A cos(t + )

Since = t =

Point Q moves with simple

harmonic motion

What is the amplitude and phase constant (relative to an x axis to the right) of the simple harmonic motion of

the ball’s shadow ?

_____ 0.50 m and 0

_____ 1.00 m and 0

_____ 0.50 m and

_____ 1.00 m and

Example 1

While riding behind a car traveling at 3.00 m/s, you

notice that one of the car’s tires has a small

hemispherical bump on its rim, as shown.

Example 2

a) Explain why the bump, from your viewpoint behind the car, executes simple harmonic motion.

b) If the radii of the car’s tires are 0.300 m, what is the bump’s period of oscillation ?

A simple pendulum exhibits periodic motion. It consists of a particle-

like bob of mass m suspended by a light

string of length L that is fixed at the upper end.

Lesson 4 : The Pendulum

Forces acting on bob

Tension in string

Gravitational force mg

Ft = mg sin always acts opposite to the displacement

of the bob

Ft is a restoring force

Ft = -mg sin = md2s

dt2

Since s = L and L is constant,

d2dt2

= -g

Lsin for small values of

d2x

dt2= -

k

mx

d2dt2

= -g

Lsin

same form

For small amplitudes ( < about 10o), the motion of a pendulum is close to simple

harmonic motion.

Instead of x = A cos(t + ),

= max cos(t + )

Instead of = k

m,

= g

L

angular frequency

for a simple pendulum

T =2

= 2L

g

= g

LT = 2

L

g

The period and frequency of a simple pendulum depend only on the length of the string and the acceleration due to gravity.

(Period and frequency are independent of mass.)

Example 1

Christian Huygens (1629-1695), the greatest clockmaker in history, suggested that an

international unit of length could be defined as the length of a simple pendulum having a

period of exactly 1s.

a) How much shorter would our length unit be had his suggestion been followed ?

b) What if Huygens had been born on another planet ? What would the value of g have to be on that planet such that the meter based on Huygen’s pendulum would have the same value as our meter ?

Example 2

A simple pendulum has a mass of 0.250 kg and a length of 1.00 m. It is displaced

through an angle of 15.0o and released. What is the

a) maximum speed

c) maximum restoring force ?

b) maximum angular acceleration

Example 3

A simple pendulum is 5.00 m long.

a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 5.00 m/s2 ?

b) What is its period if the elevator is accelerating downward at 5.00 m/s2 ?

c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 5.00 m/s2 ?

The Physical Pendulum

A hanging object that oscillates about a fixed axis that does not pass through its center of mass and the object cannot be

approximated as a point mass.

Gravitational force produces a torque about

an axis through O.

= mgd sin

Since = I,

- mgd sin = Id2dt2

- mgd sin = Id2dt2

negative sign because tends to decrease (restoring force)

If is small so that sin is almost ,

mgdd2dt2

= - ( )I

= - 2

simple harmonic motion equation

=mgd

I

Since the period of a pendulum is

T =2

,

T = 2 I

mgd

Period of a Physical Pendulum

A uniform rod of mass M and length L is pivoted about one

end and oscillates in a vertical plane. Find the period of

oscillation if the amplitude of the motion is small.

Example 4

A rigid object suspended by a wire. When the object is twisted, the twisted wire exerts a restoring torque that is proportional to the

angular position.

Torsional Pendulum

= -

is called the torsion constant of the wire

Since = Id2dt2

,

simple harmonic motion equation

d2dt2

I

= -

= I

T = 2 I

Period of a Torsional

Pendulum

Example 5

A torsional pendulum is formed by taking a meter stick of mass 2.00 kg, and attaching to its center a wire. With its upper end clamped,

the vertical wire supports the stick as the stick turns in a horizontal plane. If the resulting period is 3.00 minutes, what is the torsion

constant for the wire ?