Question

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

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

Answer 1

Answer: (D)

kinetic energy at mean position= total energy-potential energy at mean position$=9J-5J=4J$
kinetic energy at mean position$=\frac{1}{2}m{v_{max}}^2$
$\Rightarrow \frac{1}{2}m{v_{max}}^2=4J$
$\Rightarrow v_{max}=\sqrt{\frac{4\times 2}{2}}m/s=2m/s$
$\Rightarrow A\omega =A\frac{2\pi }{T}\Rightarrow A\frac{2\pi }{T}=2m/s\Rightarrow T=A\pi =0.01\pi =\frac{\pi }{100}s$