Question 243: Oscillations

Question

For a particle performing S.H.M.; the total energy is ' $n$ ' times the kinetic energy, when the displacement of a particle from mean position is $\frac{\sqrt{3}}{2} \mathrm{~A}$, where A is the amplitude of S.H.M. The value of ' $n$ ' is

Accepted Answer

We know
$$ 	ext { K.E. }=\frac{1}{2} m \omega^2\left(A^2-x^2ight) 	ext { and T.E. }=\frac{1}{2} m \omega^2 A^2 $$
Given: T.E. $=\mathrm{n} 	imes$ K.E.
$$ \begin{aligned} & \frac{1}{2} m \omega^2 A^2=n 	imes \frac{1}{2} m \omega^2\left(A^2-x^2ight) \ & \frac{1}{2} m \omega^2 A^2=n 	imes \frac{1}{2} m \omega^2 A^2\left(1-\frac{3}{4}ight) \ldots (given \,\,\,x=\frac{\sqrt{3}}{2} A)\end{aligned} $$
$$ \begin{array}{ll} \Rightarrow & 1=\mathrm{n}\left(1-\frac{3}{4}ight) \ 	herefore & \mathrm{n}=4 \end{array} $$

Answer

2

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3

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4

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6

Oscillations

For a particle performing S.H.M.; the total energy is ' n ' times the kinetic energy, when the displacement of a particle from mean position is \frac{\sqrt{3}}{2} \mathrm{~A}, where A is the amplitude of S.H.M. The value of ' n ' is

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