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
- $\frac{\pi}{100} s$
- $\frac{\pi}{50} s$
- $\frac{\pi}{20} s$
- $\frac{\pi}{10} s$
The Correct Option is A
Solution and Explanation
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
$T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \times \sqrt{\frac{2}{8 \times 10^{4}}}$
$\Rightarrow T=\frac{\pi}{100} s$
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Concepts Used:
Energy In Simple Harmonic Motion
We can note there involves a continuous interchange of potential and kinetic energy in a simple harmonic motion. The system that performs simple harmonic motion is called the harmonic oscillator.
Case 1: When the potential energy is zero, and the kinetic energy is a maximum at the equilibrium point where maximum displacement takes place.
Case 2: When the potential energy is maximum, and the kinetic energy is zero, at a maximum displacement point from the equilibrium point.
Case 3: The motion of the oscillating body has different values of potential and kinetic energy at other points.