Simple Harmonic Motion A pendulum swinging from side to side is an example of both periodic and simple harmonic motion. 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 the equilibrium position. Free-body diagram of a pendulum in its equilibrium position: T FwFw T is the tension in the string and F w is the weight of the pendulum bob. ΣF = 0T = F w

The free-body diagram at the amplitude: where F w is the weight of the pendulum bob, F r is the restoring force, T is the tension in the string, and F 1 is the other component of F w. T F1F1 FwFw FrFr θ θ The free-body diagram at the amplitude: where F w is the weight of the pendulum bob, F r is the restoring force, T is the tension in the string, and F 1 is the other component of F w. FrFr θ

In order for the restoring force (F r ) to be proportional to the displacement, 0 < θ <≈ 10°, because θ ≈ sin θ for angles less than 10°.

The Vocabulary of a Pendulum Amplitude is the maximum displacement from the 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 complete vibration (s).

Characteristics of a Pendulum In the absence of air resistance, the period of a pendulum is independent of the mass of the pendulum bob. The period of a pendulum is independent of the amplitude if 0 < θ <≈ 10°. The period of a pendulum is directly proportional to the square root of its length.

The period of a pendulum is inversely proportional to the square root of the acceleration 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/s 2 (dependent on location).

Pendulum Problems On the top of a small hill a pendulum 1.45 m long has a period of 2.47 s. What is g for this location? T = 2.47 sl = 1.45 m T = 2π(l/g) 1/2 g = 4π 2 l/T 2 = 4 × × 1.45 m/(2.47 s) 2 g = 9.38 m/s 2

Another Pendulum Problem A 0.70 kg pendulum bob at the end of a 0.85 m string is pulled back 0.90 m and released. m = 0.70 kgl = 0.85 mΔx = 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 F1F1 FwFw FrFr θ θ The free-body diagram at the amplitude: FrFr θ θ ≈ sin θ for small angles, therefore θ = sin -1 (0.10/0.80) and sin θ = F r /F w.

F r = F w × sinθ = mgsinθ F r = 0.70 kg × 9.80 m/s 2 × 0.10 m/0.85 m F r = 0.81 N a = F r /m = 0.81 N/0.70 kg = 1.2 m/s 2 (b) What is the acceleration as the pendulum passes through the equilibrium position? 0 because F r = 0.

(c) What is the net force at the instant of release? F net = F r = 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/s 2 )) 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 Questions What is the distance traveled by an object moving with simple harmonic motion during a time that is equal to its period? The distance would be 2 × Δx max where x max is the amplitude of vibration.

A pendulum is hung from the ceiling of a stationary elevator. After determining the period of the pendulum, the elevator accelerates upward, accelerates downward, and finally moves downward with a constant velocity. The period is determined while undergoing these motions.

Did the period of the pendulum change while the elevator accelerated upward? During the upward acceleration, the apparent value for g increases. Remember, during the upward acceleration, there has to be a net force acting upward in accordance with Newton’s 2 nd 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 while the 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 while the elevator moved downward with a constant velocity? While the elevator moves downward with a constant velocity, the value of g remains the same. The period of a pendulum will remain the same. In the case of free fall, the pendulum does not oscillate.