Simple Harmonic Motion A Power Point Presentation by

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.

Objectives: After finishing this unit, you should be able to: • 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 acceleration a in terms of displacement x or time t. • Describe the motion of pendulums and calculate the length required to produce a given frequency.

Periodic Motion 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, Period T, is the time for one complete oscillation. (seconds, s) Amplitude A Frequency, Frequency f, is the number of complete oscillations per second. Hertz (s-1)

Example 1: The suspended mass makes 30 complete oscillations in 15 s. What is the period and frequency of the motion? x F Period: T = 0. 500 s Frequency: f = 2. 00 Hz

Simple Harmonic Motion, SHM Simple harmonic motion is periodic motion in the absence of friction and produced by a restoring force that is directly proportional to the displacement and oppositely directed. x F A restoring force, F, acts in the direction opposite the displacement of the oscillating body. F = -kx

Hooke’s Law When a spring is stretched, there is a restoring force that is proportional to the displacement. F = -kx x m F The spring constant k is a property of the spring given by: k= DF Dx

Work Done in Stretching a Spring Work done ON the spring is positive; work BY spring is negative. x From Hooke’s law the force F is: F (x) = kx F x 1 x 2 m F To stretch spring from x 1 to x 2 , work is: (Review module on work)

Example 2: A 4 -kg mass suspended from a spring produces a displacement of 20 cm. What is the spring constant? The stretching force is the weight (W = mg) of the 4 -kg mass: 20 cm F = (4 kg)(9. 8 m/s 2) = 39. 2 N F m Now, from Hooke’s law, the force constant k of the spring is: k= DF Dx = 39. 2 N 0. 2 m k = 196 N/m

Example 2(cont. : The mass m is now stretched a distance of 8 cm and held. What is the potential energy? (k = 196 N/m) The potential energy is equal to the work done in stretching the spring: 0 U = 0. 627 J 8 cm F m

Displacement in SHM 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.

Velocity in SHM 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 -).

Acceleration in SHM +a -x +x -a 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.

Acceleration vs. Displacement x a v 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.

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/s 2 a m Note: When the displacement is +7 cm (downward), the acceleration is -14. 0 m/s 2 (upward) independent of motion direction. +x

Example 4: What is the maximum acceleration for the 2 -kg mass in the previous problem? (A = 12 cm, k = 400 N/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 m Maximum Acceleration: amax = ± 24. 0 m/s 2 +x

Conservation of Energy 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. x a v m x = -A x=0 x = +A For any two points A and B, we may write: ½mv. A 2 + ½kx. A 2 = ½mv. B 2 + ½kx. B 2

Energy of a Vibrating System: A x a v m x = -A x=0 B x = +A • At points A and B, the velocity is zero and the acceleration is a maximum. The total energy is: U + K = ½k. A 2 x = A and v = 0. • At any other point: U + K = ½mv 2 + ½kx 2

Velocity as Function of Position. x a v m x = -A vmax when x = 0: x=0 x = +A

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? ½mv 2 + ½kx 2 = ½k. A 2 m v = ± 1. 60 m/s +x

Example 5 (Cont. ): What is the maximum velocity for the previous problem? (A = 10 cm, k = 800 N/m, m = 2 kg. ) The velocity is maximum when x = 0: 0 ½mv 2 + ½kx 2 = ½k. A 2 m v = ± 2. 00 m/s +x

Period and Frequency as a Function of Mass and Spring Constant. For a vibrating body with an elastic restoring force: Recall that F = ma = -kx: -kx The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units.

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? x a v m x = -0. 2 m x=0 f = 2. 25 Hz 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) x a v m x = -0. 2 m x=0 x = +0. 2 m Acceleration is a maximum when x = A a = 40 m/s 2

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. ) x a v m v = -2 f A sin 2 f t x = -0. 2 m x = 0 x = +0. 2 m (Note: q in rads) v = -0. 916 m/s The minus sign means it is moving to the left.

Example 7: At what time will the 2 -kg mass be located 12 cm to the left of x = 0? (A = 20 cm, f = 2. 25 Hz) -0. 12 m x a v m x = -0. 2 m x = 0 t = 0. 157 s x = +0. 2 m

The Simple Pendulum The period of a simple pendulum is given by: L For small angles q. mg

Example 8. What must be the length of a simple pendulum for a clock which has a period of two seconds (tick-tock)? L L = 0. 993 m

Summary 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 m F

Summary (Cont. ) Hooke’s Law: In a spring, there is a restoring force that is proportional to the displacement. x The spring constant k is defined by: m F

Summary (SHM) x a v m x = -A x=0 x = +A Conservation of Energy: ½mv. A 2 + ½kx. A 2 = ½mv. B 2 + ½kx. B 2

Summary (SHM)

Summary: Period and Frequency for Vibrating Spring. a v x m x = -A x=0 x = +A

Summary: Simple Pendulum L

CONCLUSION: Simple Harmonic Motion