Question

The time period of a particle undergoing shm is 16 seconds it starts motion from the mean position after 2 second its velocity 0.4 m per second the amplitude is

Answer 1

Final Answer : 1.44 m

Steps :
1) Let the equation of SHM be
$y = A sin(\omega t) \\ \\ => v = A \frac{2 \pi}{T}cos( \frac{2\pi}{T}*t ) \\ \\ => 0.4 = A * \frac{2\pi}{16} * cos(\frac{2\pi }{16}*2) \\ \\ => A = 1.44 m$

Answer 2

expression of velocity in SHM,

v = Aω.cos(ωt)

Now ω = 2π / T

Given, T = 16 sec ; t = 2 sec

Thus, ωt = (2π / 16) x 2 = π / 4

Also given, v = 0.4 m per sec.

therefore,

0.4 = A.(2π/16) cos π/4

or, 0.4 = A (0.2777)

or, A = 0.4 / 0.2777

or, A = 1.44

So the amplitude is 1.44 m.

or, A = 0.4 / (0.3927)