question_answer

A particle executes SHM in a straight line. In the first second starting from rest it travels 'a' distance a and in the next second a distance 'b' in the same direction. The amplitude of S.H.M will be

A) \[\frac{2{{a}^{2}}}{3a-b}\]

B) \[a-b\]

C) \[2a-b\]

D) \[a/b\]

Correct Answer: A

Solution :

\[x=A\cos \,wt\]  (as it starts from rest at t=0)

\[A-a=A\cos w\]                         ...(1)

\[A-(a+b)=A\cos 2w\]                 ...(2)

Solving (1) and (2) for A we get \[A=\frac{2{{a}^{2}}}{3a-b}\]