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प्रश्न
Draw a graph to show the variation of P.E., K.E. and total energy of a simple harmonic oscillator with displacement.
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उत्तर
The potential energy (PE) of a simple harmonic oscillator is = `1/2 kx^2`
= `1/2 mω^2x^2` .....(i)
Where, k = force constant = `mω^2`
When PE is plotted against displacement x, we will obtain a parabola.
When x = 0, PE = 0
When x = ± A, PE = maximum = `1/2 mω^2A^2`
KE of a simple harmonic oscillator = `1/2 mv^2` .....`[∵ v = ωsqrt(A^2 - x^2)]`
= `1/2 m[ωsqrt(A^2 - x^2)]^2`
= `1/2 mω^2(A^2 - x^2)` ......(ii)
This is also a parabola if plotting KE against displacement x.
i.e., KE = 0 at x = ± A
And KE = `1/2 mω^2A^2` at x = 0
Now, the total energy of the simple harmonic oscillator = PE + KE .......[Using equations (i) and (ii)]
= `1/2 mω^2x^2 + 1/2 mω^2 (A^2 - x^2)`
= `1/2 mω^2x^2 + 1/2 mω^2A^2 - 1/2 mω^2x^2`
TE = `1/2 mω^2A^2`
Which is constant and does not depend on x.
Plotting under the above guidelines KE, PE and TE versus displacement x-graph as follows
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