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.