Question

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

• A. $\frac{\pi }{100}$ sec
• B. $\frac{\pi }{50}$ sec
• C. $\frac{\pi }{20}$ sec
• D. $\frac{\pi }{10}$ sec

Answer 1

Answer: (A)

$\frac{\pi }{100}$ sec

$\text{Given:-}$ Potential energy = 5 KJ

Total energy = 9 KJ

$\text{To find:-}$ Time period

$\text{Solution:-}$

Kinetic energy = Total energy - potential energy

$=9-5=4$ KJ At mean position

We know that

$\frac{1}{2}m{\omega }^2{A}^2=Kineticenergy$

$\frac{1}{2}\times 2\times {\omega }^2\times ({10}^{-2}{)}^2=4$

${\omega }^2\times {10}^{-4}=4$

${\omega }^2=4\times {10}^{-4}$

$\omega =2\times {10}^2$

$T=\frac{2\pi }{\omega }=\frac{2\pi }{2\times {10}^2}$

= $\frac{\pi }{100}sec$

$\text{Hence the correct option is A}$