• # 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) $a/b$

 $x=A\cos \,wt$  (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 $A=\frac{2{{a}^{2}}}{3a-b}$