Question

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

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

Answer 1

Answer: (D)

$P.E.=5J, \, T.E.=9J$
Using mechanical energy conservation
$T.E. \, =P.E. \, +K.E.$
$9=5+K.E.$
$\Rightarrow K.E.=4$ (Maximum)
$K.E._{m a x}$ at mean position $=\frac{1}{2}\omega ^{2}A^{2}m$
$4=\frac{1}{2}\times 2\times \left(\frac{2 \pi }{t}\right)^{2}\times \frac{1}{100}\times \frac{1}{100}$
$T^{2}=\frac{\pi ^{2}}{100 \times 100}$
$T=\frac{\pi }{100}s$