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Chapter 14 - Simple Harmonic Motion

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Chapter 14 - Simple Harmonic Motion

A PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

© 2007

Photo by Mark Tippens

A TRAMPOLINE exerts a restoring force on the jumper that is directly proportional to the average force required to displace the mat. Such restoring forces provide the driving forces necessary for objects that oscillate with simple harmonic motion.

- Write and apply Hooke’s Law for objects moving with simple harmonic motion.

- Write and apply formulas for finding the frequency f,period T,velocity v, or accelerationa in terms of displacementxor time t.

- Describe the motion of pendulums and calculate the length required to produce a given frequency.

AmplitudeA

Simple periodic motion is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time.

Period, T, is the time for one complete oscillation. (seconds,s)

Frequency, f, is the number of complete oscillations per second. Hertz (s-1)

x

F

Period: T = 0.500 s

Frequency: f = 2.00 Hz

x

F

Simple harmonic motionis periodic motion in the absence of friction and produced by a restoring force that is directly proportional to the displacement and oppositely directed.

A restoring force, F, actsin the direction opposite the displacement of the oscillating body.

F = -kx

x

F

m

DF

Dx

k =

When a spring is stretched, there is a restoring force that is proportional to the displacement.

F = -kx

The spring constant k is a property of the spring given by:

x

F

m

F (x) = kx

F

x1

x2

Work done ONthe spring is positive; work BY spring is negative.

From Hooke’s law the force F is:

To stretch spring from x1 to x2 , work is:

(Review module on work)

20 cm

F

m

DF

Dx

39.2 N

0.2 m

k = =

The stretching force is the weight (W = mg) of the 4-kg mass:

F = (4 kg)(9.8 m/s2) = 39.2 N

Now, from Hooke’s law, the force constant k of the spring is:

k = 196 N/m

8 cm

F

0

m

The potential energy is equal to the work done in stretching the spring:

U = 0.627 J

x

m

x = -A

x = 0

x = +A

- Displacement is positive when the position is to the right of the equilibrium position (x = 0) and negative when located to the left.

- The maximum displacement is called the amplitude A.

v (-)

v (+)

m

x = -A

x = 0

x = +A

- Velocity is positive when moving to the right and negative when moving to the left.

- It is zero at the end points and a maximum at the midpoint in either direction (+ or -).

+a

-a

-x

+x

m

x = -A

x = 0

x = +A

- Acceleration is in the direction of the restoring force. (a is positive when x is negative, and negative when x is positive.)

- Acceleration is a maximum at the end points and it is zero at the center of oscillation.

a

v

x

m

x = -A

x = 0

x = +A

Given the spring constant, the displacement, and the mass, the acceleration can be found from:

or

Note: Acceleration is always opposite to displacement.

a

+x

m

Example 3:A 2-kg mass hangs at the end of a spring whose constant is k = 400 N/m. The mass is displaced a distance of 12 cm and released. What is the acceleration at the instant the displacement is x = +7 cm?

a = -14.0 m/s2

Note: When the displacement is +7 cm (downward), the acceleration is -14.0 m/s2 (upward) independent of motion direction.

+x

m

The maximum acceleration occurs when the restoring force is a maximum; i.e., when the stretch or compression of the spring is largest.

F = ma = -kx

xmax = A

Maximum Acceleration:

amax = ± 24.0 m/s2

a

v

x

m

x = -A

x = 0

x = +A

The total mechanical energy(U + K) of a vibrating system is constant; i.e., it is the same at any point in the oscillating path.

For any two points A and B, we may write:

½mvA2 + ½kxA2 = ½mvB2 + ½kxB2

m

a

v

x

x = -A

x = 0

x = +A

A

B

- At points A and B, the velocity is zero and the acceleration is a maximum. The total energy is:

U + K = ½kA2x = A and v = 0.

- At any other point: U + K = ½mv2 + ½kx2

m

a

v

x

x = -A

x = 0

x = +A

vmaxwhen x = 0:

+x

m

Example 5:A 2-kg mass hangs at the end of a spring whose constant is k = 800 N/m. The mass is displaced a distance of 10 cm and released. What is the velocity at the instant the displacement is x = +6 cm?

½mv2 + ½kx 2 = ½kA2

v = ±1.60 m/s

0

+x

m

The velocity is maximum when x = 0:

½mv2 + ½kx 2 = ½kA2

v = ± 2.00 m/s

w = 2f

The reference circle compares the circular motion of an object with its horizontal projection.

x = Horizontal displacement.

A = Amplitude (xmax).

q = Reference angle.

The velocity (v) of an oscillating body at any instant is the horizontal component of its tangential velocity (vT).

vT = wR = wA; w = 2f

v = -vTsin ; = wt

v = -wA sin w t

v = -2f A sin 2ft

The acceleration(a) of an oscillating body at any instant is the horizontal component of its centripetal acceleration (ac).

a = -accos q = -ac cos(wt)

R = A

a = -w2A cos(wt)

For any body undergoing simple harmonic motion:

Since a = -4p2f2x and T = 1/f

The frequency and the period can be found if the displacement and acceleration are known. Note that the signs of aand x will always be opposite.

For a vibrating body with an elastic restoring force:

Recall that F = ma = -kx:

The frequencyfand the periodTcan be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units.

m

a

v

x

x = 0

x = -0.2 m

x = +0.2 m

Example 6:The frictionless system shown below has a 2-kg mass attached to a spring (k = 400 N/m). The mass is displaced a distance of 20 cm to the right and released.What is the frequency of the motion?

f = 2.25 Hz

m

a

v

x

x = 0

x = -0.2 m

x = +0.2 m

Example 6 (Cont.):Suppose the 2-kg mass of the previous problem is displaced 20 cm and released (k = 400 N/m). What is the maximum acceleration? (f = 2.25 Hz)

Acceleration is a maximum when x = A

a = 40 m/s2

m

a

v

x

x = 0

x = -0.2 m

x = +0.2 m

Example 6:The 2-kg mass of the previous example is displaced initially at x = 20 cm and released. What is the velocity 2.69 s after release? (Recall that f = 2.25 Hz.)

v = -2f A sin 2ft

(Note: q in rads)

The minus sign means it is moving to the left.

v = -0.916 m/s

-0.12 m

m

a

v

x

x = 0

x = -0.2 m

x = +0.2 m

t = 0.157 s

L

mg

The period of a simple pendulum is given by:

For small angles q.

L

L = 0.993 m

The period T of a torsion pendulum is given by:

Where k’ is a torsion constant that depends on the material from which the rod is made; I is the rotational inertia of the vibrating system.

Example 9:A 160 g solid disk is attached to the end of a wire, then twisted at 0.8 rad and released. The torsion constant k’ is 0.025 N m/rad. Find the period.

(Neglect the torsion in the wire)

For Disk:

I = ½mR2

I = ½(0.16 kg)(0.12 m)2 =0.00115 kg m2

T = 1.35 s

Note: Period is independent of angular displacement.

x

F

m

Simple harmonic motion (SHM) is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time.

The frequency (rev/s) is the reciprocal of the period (time for one revolution).

x

F

m

Hooke’s Law: In a spring, there is a restoringforce that is proportional to the displacement.

The spring constant k is defined by:

m

a

v

x

x = -A

x = 0

x = +A

Conservation of Energy:

½mvA2 + ½kxA2 = ½mvB2 + ½kxB2

m

a

v

x

x = -A

x = 0

x = +A

L