Question

56. A bob executes SHM of period 16s, two seconds after it passes through its centre of
oscillations, its velocity is found to be 0.05ms-1. Find the amplitude. Ans: 0.18m​

Answer 1

Answer:

Amplitude = 0.18 m

Explanation:

T = 16 sec.

t = 2 sec.

v = 0.05 m/s

W(angular frequency) = (2 x pie) / T

v = Awsinwt