Question

For a body is simple harmonic motion velocity at mean position is 4m/s if the time period is 3.14s ,find its amplitude​

Answer 1

Given info : for a body is simple harmonic motion, velocity at mean position is 4m/s if the time period is 3.14 sec.

To find : the amplitude of the shm is ...

solution : we know, velocity of the particle in simple harmonic motion is given by, v = $\omega\sqrt{A^2-x^2}$

where ω is angular frequency , A is amplitude and x is the position of the particle.

at mean position, x = 0

we know, angular frequency, ω = $\frac{2\pi}{T}$ , T is time period.

∴ v = $\frac{2\pi}{T}\times A$

⇒ 4 m/s = $\frac{2\times3.14}{3.14 s}\times A$

⇒ A = 2m

therefore the amplitude of the simple harmonic motion is 2 m.