Question

The equation for the displacement of a particle executing SHM is x = 5sin (2pit)cm Then the velocity at 3cm from the mean position is( in cm//s)

Answer 1

Answer:

ω = 2π rad/sec A = 5cm x = 3cm v = ω√(A2-x^2) = 2π × √(25 - 9) = 8π cm/s

Answer 2

Answer:

Here is step by step

Explanation:

Given, x=3cm

x=5sin(2$\pi$t)

compare this with x=Asinωt

A=5cm

ω=2$\pi$

we have,  

v=ω$\sqrt{A^{2}-\\x^{2} }$

v=2$\pi$$\sqrt{5^{2} -3^{2} }$

v=2$\pi$$\sqrt{16}$

v=2$\pi$4

v=8$\pi$

HOPE IT HELPS YOU