Find the displacement of a simple harmonic os | vedclass

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Question

(B) Let the amplitude of the oscillator be $A$ and the displacement at time $t$ be $x$.The total energy $E$ of the simple harmonic oscillator is given

Vedclass · Accepted Answer

$x = \pm A/\sqrt{2}$

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(B) Let the amplitude of the oscillator be $A$ and the displacement at time $t$ be $x$.The total energy $E$ of the simple harmonic oscillator is given by $E = \frac{1}{2} k A^{2}$.The potential energy $U$ at displacement $x$ is given by $U = \frac{1}{2} k x^{2}$.According to the problem,the potential energy is half of the maximum energy,so $U = \frac{1}{2} E$.Substituting the expressions for $U$ and $E$,we get $\frac{1}{2} k x^{2} = \frac{1}{2} (\frac{1}{2} k A^{2})$.Simplifying the equation,we get $x^{2} = \frac{A^{2}}{2}$.Taking the square root on both sides,we find $x = \pm \frac{A}{\sqrt{2}}$.The $\pm$ sign indicates that the oscillator can be at this displacement on either side of the mean position.

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

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