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Question

Physics Question on simple harmonic motion

An SHM is given by y=5[sin(3πt)+3cos(3πt)]y = 5 [\sin (3 \pi t) + \sqrt{3} \cos (3 \pi t)] What is the amplitude of the motion of yy in metre?

A

10

B

20

C

1

D

5

Answer

10

Explanation

Solution

Given equation is y=5[sin3πt+3cos3πt]y=5[\sin 3 \pi t+\sqrt{3 \cos 3 \pi t}] or y=5×2[sin3πt×12=32cos3πt]y=5 \times 2\left[\sin 3 \pi t \times \frac{1}{2}=\frac{\sqrt{3}}{2} \cos 3 \pi t\right] or y=10sin(3πtπ3)y=10 \sin \left(3 \pi t \frac{\pi}{3}\right)