Question:
If $x, v$ and $a$ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $T$, then, which of the following does not change with time?
If $x, v$ and $a$ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $T$, then, which of the following does not change with time?
Updated On: Jul 28, 2022
- $a^2 T^2 +4 \pi^2 v^2$
- $\frac{a T}{x}$
- $aT + 2 \pi v$
- $\frac{a T}{v}$
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The Correct Option is B
Solution and Explanation
For a particle executing simple harmonic motion,
acceleration, $a = - \omega^2 x$ where $\omega = \frac{2 \pi}{T}$
$ a = - \frac{4 \pi^2}{T^2} x $ or $\frac{aT}{x} = - \frac{4 \pi^2}{T}$
The period of oscillation T is a constant.
$ \therefore \:\: \frac{aT}{x} $ is a constant.
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