Question 237: Oscillations

Question

A particle is executing S.H.M. of amplitude ' $A$ '. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is

Accepted Answer

The potential energy in simple harmonic motion is given by $P = \frac{1}{2}kx^2$.
The maximum potential energy happens when the particle is at the farthest point from the center. This value is $P_{\max} = \frac{1}{2} k A^2$.
We are told that the current potential energy is half of this maximum value. So,

$$\frac{1}{2} k x^2 = \frac{1}{2} \left( \frac{1}{2} k A^2 \right)$$
If we simplify both sides, we get:

$$x^2 = \frac{A^2}{2}$$

So the displacement from the center is:

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

Answer

$\pm \frac{\mathrm{A}}{4}$

Answer

$\pm \frac{A}{2}$

Answer

$\pm \frac{\mathrm{A}}{\sqrt{3}}$

Answer

$$ \pm \frac{A}{\sqrt{2}} $$

Oscillations

A particle is executing S.H.M. of amplitude ' A '. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is

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