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Simple Harmonic Motion Lesson 2 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 F m To stretch spring from x1 to x2 , work is: Work ½kx ½kx 2 2 x1 x2 2 1 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/s2) = 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: Work ½kx ½kx 2 2 8 cm 0 m 2 1 U ½kx ½(196 N/m)(0.08 m) 2 U = 0.627 J F 2 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.) F ma kx • Acceleration is a maximum at the end points and it is zero at the center of oscillation. Acceleration vs. Displacement 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: F ma kx or kx a m 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? kx a m (400 N/m)(+0.07 m) a 2 kg a = -14.0 m/s2 a m Note: When the displacement is +7 cm (downward), the acceleration is -14.0 m/s2 (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 kA 400 N( 0.12 m) a m 2 kg Maximum Acceleration: m amax = ± 24.0 m/s2 +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. a v x m x = -A x=0 x = +A For any two points A and B, we may write: ½mvA2 + ½kxA 2 = ½mvB2 + ½kxB 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 = ½kA2 x = A and v = 0. • At any other point: U + K = ½mv2 + ½kx2 Velocity as Function of Position. x a v m x = -A 1 2 x=0 k v A2 x 2 m mv kx kA 2 1 2 2 1 2 vmax when x = 0: x = +A 2 v k A 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 k v A2 x 2 m 800 N/m v (0.1 m) 2 (0.06 m) 2 2 kg v = ±1.60 m/s m +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 ½mv2 + ½kx 2 = ½kA2 v k 800 N/m A (0.1 m) m 2 kg v = ± 2.00 m/s m +x The Simple Pendulum The period of a simple pendulum is given by: L T 2 g L For small angles q. 1 f 2 g L 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 T 2 g 2 2 L T 4 ; g (2 s) 2 (9.8 m/s 2 ) L 2 4 L T 2g L= 2 4 L = 0.993 m To Be Continued…