Question

The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5 J. If its totalenergy is 9J and its amplitude is 0.01 m, its time period would be

• A. $\pi$/10 sec
• B. $\pi$/20 sec
• C. $\pi$/50 sec
• D. $\pi$/100 sec

Answer 1

The correct option is D $\pi$/100 sec

As minimum melocity in SHM occurs at centre 0i mean position, $\therefore$ KE at centre = Total energy - P.E at centre = 9 - 5 = 4 joules.

$\begin{array}{c}K.E=1/2M{V}^2\\ 4=1/2.2\left(V^2\right)\\ V^2=4,V=2m/s\\ InSHM,Vman=AW,A=0.01m\\ 2=0.01\times W,W=200\\ Timeperiod\left(T\right)=\frac{2\pi }{W}=\frac{2\pi }{200}=\frac{\pi }{100}\sec .\end{array}$