Simple Harmonic MotionSimple Harmonic MotionA pendulum swinging from side to side is

anexample of both periodic and simple

harmonicmotion.

Periodic motion is when an object continually

moves back and forth over a definite path in

equal intervals of time.

Simple harmonic motion (SHM) is when the

acceleration of an object is proportional to the

displacement of the object from its equilibrium

position and is always directed toward theequilibrium position.

Free-body diagram of a pendulum in itsequilibrium position:•

T

Fw

T is the tension in the string and Fwis the weight of the pendulum bob.

ΣF = 0 T = Fw

The free-body diagram at the amplitude:

where Fw is the weight of the pendulum bob, Fr

is the restoring force, T is the tension in thestring, and F1 is the other component of Fw.

T

F1

Fw

Fr θθ •

The free-body diagram at the amplitude:

where Fw is the weight of the pendulum bob, Fr

is the restoring force, T is the tension in thestring, and F1 is the other component of Fw.

Frθ •

In order for the restoring force (Fr) to beproportional to the displacement, 0 < θ

<≈ 10°,because θ ≈ sin θ for angles less than

10°.

The Vocabulary of a PendulumThe Vocabulary of a PendulumAmplitude is the maximum displacement

fromthe equilibrium position (m).

Frequency is the number of complete vibrations

usually given in /s or hz (1 hz = 1 cycle/s).

Period is the time needed for one completevibration (s).

Characteristics of a PendulumCharacteristics of a PendulumIn the absence of air resistance, the period

of apendulum is independent of the mass of thependulum bob.

The period of a pendulum is independent of the

amplitude if 0 < θ <≈ 10°.

The period of a pendulum is directlyproportional to the square root of its length.

The period of a pendulum is inverselyproportional to the square root of theacceleration due to gravity.

The period of a pendulum is given by

T = 2π(l/g)1/2

where T is the period usually in s, l is the length

usually in m, and g = 9.80 m/s2 (dependent on

location).

Pendulum ProblemsPendulum ProblemsOn the top of a small hill a pendulum 1.45 mlong has a period of 2.47 s. What is g for thislocation?

T = 2.47 s l = 1.45 m

T = 2π(l/g)1/2

g = 4π2l/T2 = 4 × 3.142 × 1.45 m/(2.47 s)2

g = 9.38 m/s2

Another Pendulum ProblemAnother Pendulum ProblemA 0.70 kg pendulum bob at the end of a

0.85 mstring is pulled back 0.90 m and released.

m = 0.70 kg l = 0.85 m Δx = 0.10 m

(a) What is the acceleration of the pendulum

bob at the instant it is released?

The free-body diagram at the amplitude:

T

F1

Fw

Fr θθ •

The free-body diagram at the amplitude:

Frθ •

θ ≈ sin θ for small angles, thereforeθ = sin-1(0.10/0.80) and sin θ = Fr/Fw.

Fr = Fw × sinθ = mgsinθ

Fr = 0.70 kg × 9.80 m/s2 × 0.10 m/0.85 m

Fr = 0.81 N

a = Fr/m = 0.81 N/0.70 kg = 1.2 m/s2

(b) What is the acceleration as the pendulum passes through the equilibrium position?

0 because Fr = 0.

(c) What is the net force at the instant of release?

Fnet = Fr = 0.81 N (See part (a)).

(d) What is the period of the pendulum?

T = 2π(l/g)1/2

T = 2 × 3.14 × ((0.85 m/9.80 m/s2))1/2

T = 1.9 s

(e) What is the period of the pendulum if it was

in a freely falling elevator?

T = 2π(l/g)1/2

In free fall, the “effective g” equals zero so

the period would be infinite which correlates

to no swing.

Wrap Up QuestionsWrap Up QuestionsWhat is the distance traveled by an objectmoving with simple harmonic motion

during atime that is equal to its period?

The distance would be 2 × Δxmax where xmax is

the amplitude of vibration.

A pendulum is hung from the ceiling of astationary elevator. After determining theperiod of the pendulum, the elevatoraccelerates upward, accelerates

downward,and finally moves downward with a

constantvelocity. The period is determined whileundergoing these motions.

Did the period of the pendulum change whilethe elevator accelerated upward?

During the upward acceleration, the apparentvalue for g increases. Remember, during theupward acceleration, there has to be a net

forceacting upward in accordance with Newton’s 2nd

Law.

The period of a pendulum is given by:

T = 2 × π × (l/g)1/2.

An increase in g will decreases the period, T.

Did the period of the pendulum change whilethe elevator accelerated downward?

During the downward acceleration, the apparent

value for g decreases.

The period of a pendulum is given by:

T = 2 × π × (l/g)1/2.

A decrease in g will increases the period, T.

Did the period of the pendulum change whilethe elevator moved downward with a constantvelocity?

While the elevator moves downward with aconstant velocity, the value of g remains thesame.

The period of a pendulum will remain the same.

In the case of free fall, the pendulum does notoscillate.