Question

Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by x = ${\displaystyle 5\sin \left(\frac{\pi t}{3}\right)}$m.

Answer 1

Given:

x = ${\displaystyle 5\sin \left(\frac{\pi t}{3}\right)}$m

t = 1s

∴ ν = ${\displaystyle \frac{dx}{dt}=\frac{d}{dt}\left(5\sin \left(\frac{\pi t}{3}\right)\right)=5\cos \left(\frac{\pi t}{3}\right)\times \frac{\pi }{3}}$

Put t = 1s ...(Given)

∴ ν = ${\displaystyle 5\cos \left(\frac{\pi }{3}\right)\times \frac{\pi }{3}}$ = 2.6179 m/s

Answer 2

t = 1s, x = 5 sin60

A = 5, x = ${\displaystyle 2.5\sqrt{3}}$ ...........(given)

v = ${\displaystyle w\sqrt{A^2-x^2}}$ .........(Formula)

v = ${\displaystyle \frac{\pi }{3}\sqrt{25-18.74}}$

v = 1.04 × 2.5

v = 2.61 m/s