Question

A particle having potential energy 1/3 of the maximum value of a distant of 4 cm from a mean position. The amplitude of the motion is:

• A. $4\sqrt{3}$
• B. $6\sqrt{3}$
• C. $2\sqrt{6}$
• D. $2/\sqrt{6}$

Answer 1

Answer: (A)

The maximum energy during S.H.M. $E_{\max }=\frac{1}{2}m{\omega }^2{A}^2$ ?(i) where A = amplitude of the S.H.M. now at x = 4 $P.E.=\frac{E_{\max }}{3}=\frac{1}{2}m{\omega }^2{x}^2$ ?(ii) Then from equation (i) and (ii) we get $\frac{1}{6}m{\omega }^2{A}^2=\frac{1}{2}m{\omega }^2{x}^2$ $x^2=\frac{A^2}{3},$ so $A=\sqrt{3}x=4\sqrt{3}$