# Answer on Mechanics | Relativity Question for aakriti goel

Question #1614

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.

(i)displacement.

(2) velocity (3) acceleration after 1/3 sec starting from mean position.

Expert's answer

(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 π

(ii) then the velosity would be v = 5π cos (π t) [cm/s] v(t=1/3) =2.5π

(iii) acceleration a = -5 π

^{2}sin (π t) a(t=1/3)= -2.5π^{2}√3Need a fast expert's response?

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