Simple Harmonic Motion Lesson 2

Work Done in Stretching a Spring Work done ON the spring is positive; work BY spring is negative. F x m From Hooke’s law the force F is: F (x) = kx x1 x2 F To stretch spring from x1 to x2 , work is:

Example 2: A 4-kg mass suspended from a spring produces a displacement of 20 cm. What is the spring constant? F 20 cm m 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 = = 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) F 8 cm m The potential energy is equal to the work done in stretching the spring: U = 0.627 J

Displacement in SHM m x = 0 x = +A x = -A x 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 m v (-) v (+) x = 0 x = +A 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 m -x +x x = +A 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 v a m x = 0 x = +A 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? m +x a 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.

Maximum Acceleration: 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. m +x F = ma = -kx xmax =  A Maximum Acceleration: amax = ± 24.0 m/s2

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 v a m x = 0 x = +A x = -A For any two points A and B, we may write: ½mvA2 + ½kxA 2 = ½mvB2 + ½kxB 2

Energy of a Vibrating System: x = 0 x = +A x = -A x v a A B At points A and B, the velocity is zero and the acceleration is a maximum. The total energy is: U + K = ½kA2 x =  A and v = 0. At any other point: U + K = ½mv2 + ½kx2

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

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? m +x ½mv2 + ½kx 2 = ½kA2 v = ±1.60 m/s

The velocity is maximum when x = 0: 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: m +x ½mv2 + ½kx 2 = ½kA2 v = ± 2.00 m/s

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

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

To Be Continued…