Ncert Exercises & Solutions

Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.

Hint: The maximum energy of the oscillator is equal to the potential energy at the extreme position.

Let us assume that the required displacement is x.

Step 1: Find the potential energy at the general position.

$∴$ The potential energy of the simple harmonic oscillator

$= \frac{1}{2} k x^{2}$

Step 2: Find the total energy of the oscillator.

Where k=force constant

$= m ω^{2}$

$∴ P E = \frac{1}{2} m ω^{2} x^{2} . . . (i)$

Maximum energy of the oscillator,

$T E = \frac{1}{2} m ω^{2} A^{2} [∵ x_{m a x} = A] . . . (i i)$

where A=amplitude of motion

Step 3: Find the value of the position.

Given,

$P E = \frac{1}{2} T E$

$\Rightarrow \frac{1}{2} m ω^{2} x^{2} = \frac{1}{2} [\frac{1}{2} m ω^{2} A^{2}]$

$\Rightarrow x^{2} = \frac{A^{2}}{2}$

$x = \sqrt{\frac{A^{2}}{2}} = \pm \frac{A}{\sqrt{2}}$

The sign

$\pm$ indicates either side of the mean position.