The frequency of a simple harmonic motion is 2.6 times 10^-4 s^-1. The oscillati

Question

The frequency of a simple harmonic motion is 2.6 times 10^-4 s^-1. The oscillation starts (t = 0) when the displacement has its maximum positive value of 6.5 times 10^-3 cm. The earliest possible time at which the particle can be found at x = -2.6 times 10^-3 cm is 7.1 times 10^-6 S 1.2 times 10^-5 S 1.1 times 10^-4 S A body moves simple harmonic motion according to the equation x = (2/pi)sin(4pi + pi/3). where the units are SI. At t = 2 s, the speed of the body is 13 m/s 1/__ m/s Squareroot3/pi 4 m/s 4 Squareroot3 m/s Use the following to answer questions 3-5 The object in the diagram is in circular motion. Its position at t = 0 was (A, 0). Its frequency is f. The y component of its position is given by y = y_0 + v_0y t + 1/2at^2 y = A cos 2 pi ft y = A sin ft y = A sin 2 pi ft The object in the diagram is in circular motion with frequency f. At t = 0 it was at (A, 0). The y component of its velocity is given by v_y 2 = v_0y 2 + 2a(y - y_0) v_y = 2pifA cos 2pift v_y = A sin ft v_y = 2pifA sin 2pift v_y = A cos ft

Explanation / Answer

1) x = A cos2pift

t = 1/2pifcos^-1(x/A)

2) v = dx/dt = (2/pi)*(4pi)cos(4pit+pi/3)

at t = 2 sec

v = 4 m/s

3) y = Asin(2pift)

4) vy = dy/dt = A2pifcos(2pift)

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