question_answer

A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second it travels a distance b in the same direction. The amplitude of the SHM is     

A) \[a-b\]

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

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

D) None of these

Correct Answer: C

Solution :

[c] When particle starts from mean position, \[x=A\,\cos \,\omega t\] \[\therefore \,\,\,(A-a)=A\,\cos (\omega \times 1)\] Or  \[(A-a)=A\,cos\,\omega \]             ?..(i) \[Also\,\,\,A-(a+b)=A\,\cos (\omega \times 2)\] Or   \[A-(a+b)=A\,\cos \,2\omega \]   ??(ii) After simplifying above equations, we get \[A=\left[ \frac{2{{a}^{2}}}{3a-b} \right]\]