Question

A particle of mass 0.50 kg executes a simple harmonic motion under a forve F = -(50 n//m)x. If crosses the centre of oscillation with a speed of 10 m//s, find the amplitude of the motion.

Answer 1

Answer:

A particle executing S.H.M acted upon by a force

F = -(50N/m)x

it mess m= 0.5kg

so, a = F/m = acceleration

⇒a=−100x

⇒a+ω

2

x=0

ω = angular frequency =

100

=10rad/s

velocity at mean position

ωA=10/s A = amplitude of oscillation

⇒A=

10

10

=1m

I hope it is helpful to you