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