Answer to Question #1614 in Mechanics | Relativity for aakriti goel

Question

A particle executes SHM with a time period of 2 sec and amplitude 5cm find
(i)displacement.
(2) velocity (3) acceleration after 1/3 sec starting from mean position.

Answer 1

(i) displasement is x = 5 sin (2π/T t ) = 5 sin (π t ) [cm] x(t=1/3) = 5 sin (π/3) = 5*√3/2.
(ii) then the velosity would be v = 5π cos (π t) [cm/s] v(t=1/3) =2.5π
(iii) acceleration a = -5 π² sin (π t) a(t=1/3)= -2.5π²√3