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Karnataka Board PUCPUC Science Class 11

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

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Question

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

Short/Brief Note
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Solution

Let us assume that the required displacement is x.

∴ The potential energy of the simple harmonic oscillator = `1/2 kx^2`

Where, k = force constant = `mω^2`

∴ P.E. = `1/2 mω^2x^2`  ......(i)

The maximum energy of the oscillator

TE = `1/2 mω^2A^2`  [∵ xmax = A]  ......(ii)

Where, A = Amplitude of motion

Given, P.E. = `1/2` TE

⇒ `1/2 mω^2x^2 = 1/2 [1/2 mω^2A^2]`

⇒ `x^2 = A^2/2`

or `x = sqrt(A^2/2) = +- A/sqrt(2)`

Sign ± indicates either side of mean position.

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Energy in Simple Harmonic Motion
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Chapter 14: Oscillations - Exercises [Page 103]

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NCERT Exemplar Physics Exemplar [English] Class 11
Chapter 14 Oscillations
Exercises | Q 14.31 | Page 103

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