Question

A simple harmonic oscillator of mass 0.400 kg oscillates with frequency 1.50 Hz. At t0, the oscillator is at position x 4.00 cm and is moving right with speed 42.0 cm/s a) Find the amplitude and phase constant for the oscillator. b) Write the equation for displacement of the oscillator (with numbers) c) Find the position, velocity, and acceleration at t 3.00 s. di Find the first tw o times the oscillation has position x -2 .75 cm.

Answer 1

Given simple harmonic oscillator m = 0.4 Kg f = 1.5 HZ: w = 2 pi f = 3 pi at t = 0, x_0 = 4 cm and x� = 42 cm/s a) let amplitude = A, phase constant = phi then let equation of motion be x = A sin (wt + phi) w = 2 pi f rightarrow x = A sin (2 pi f t + phi) at t = 0 x = x_0 = A sin (phi) rightarrow 4 = A sin (phi) also x� = 2 pi f A cos (2 pi f t + phi) 42 = 2 pi times 1.5 times A (cos phi) hence 42^2/4 pi^2 times 1.5^2 + 16 = A^2 rightarrow A = 5.9882 cm and phi = 0.731 rad b) x = 5.9882 cm sin (3 pi t + 0.731 rad) c) at 3s x = -3.9978cm x� = 5.9882 times 3 pi times cos (3 pi t + 0.731) = -42.018 cm/s x�� = 9 pi^2 times 5.9882 times sin (3 pi t + 0.731) = -355.11 cm/s^2 d) x = 2.75 cm = 5.9882 sin (3 pi t + 0.731) t_1 = 0.6397 s t_2 = 1.30639 s