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 amplitude is $1\, cm$. then its time period is

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

Answer 1

Answer: (A)

$\frac{\pi}{100} s$

Answer 2

Given, total energy $=9\, J$
PE at mean position $=5\, J$
So, maximum $KE =9\, J -5\, J =4\, J$
Now, in SHM
Maximum (at mean)
$KE =$ Maximum PE (at extremes)
$\therefore \frac{1}{2} k a^{2}=4\, J$
$\Rightarrow k=\frac{8}{a^{2}}=\frac{8}{10^{-4}}=8 \times 10^{4} J / m ^{2}$
Now, time period
$T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \times \sqrt{\frac{2}{8 \times 10^{4}}}$
$\Rightarrow T=\frac{\pi}{100} s$