Question

A particle is executing SHM along the x-axis given by x = A sin wt. What is the magnitude of the
average acceleration of the particle between t = 0 and t = (T/4) s, where T is the time period a
oscillation
(A) 2w²4/π
(B) wA/π
10 4w²A/π
(D) none of these​

Answer 1

given x=a sin wt . by differentiating x with respect to t we will get velocity and it will be V= aw cos wt.

now , average acceleration means change in velocity over a specified period of time.

so, average acceleration is change in velocity from time 0 second to T/4 second.

change in velocity = aw cos w(T/4)- aw cos w0

= aw cos w(π/2w) - aw

(since cos 0 is 1)

= aw cos π/2- aw

= 0 - aw ( since cos π/2 is 0)

therefore, change in velocity = - aw

now , average acceleration = - aw /( π/2w) = - 2Aw^2 / π .