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A particle executes $SHM$ with a time period of $16\,s$ . At time $t = 2\,s$ , the particle crosses the mean position while at $t = 4\,s$ , its velocity is $4\,m{s^{ - 1}}$ . The amplitude of motion in meters?
(A) $\sqrt 2 \pi $
(B) $16\sqrt 2 \pi $
(C) $\dfrac{32\sqrt 2}{\pi}$
(D) $\dfrac{4}{\pi}$

seo-qna
Last updated date: 20th Jun 2024
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Answer
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Hint: Use the formula of the simple harmonic motion, substitute it with the values at the mean position to find the phase shift. Again use the same formula and substitute the condition of the time taken and phase shift. Differentiate it with respect to time and simplifying it provides the amplitude.

Formula used:
The formula of the amplitude of the simple harmonic motion is given by
$x = \alpha \sin \left( {\dfrac{{2\pi }}{T}t + \phi } \right)$
Where $x$ is the displacement of the particle, $\alpha $ is its amplitude, $T$ is the total time taken for its simple harmonic motion and $t$ is the time taken for the particular position.

Complete step by step solution:
It is given that
Time taken for the simple harmonic motion, $T = 16\,s$
While crossing the mean position, $t = 2s$
At the time taken of $t = 4\,s$ , velocity, $v = 4\,m{s^{ - 1}}$
Using the formula of the simple harmonic motion,
$x = \alpha \sin \left( {\dfrac{{2\pi }}{T}t + \phi } \right)$
Substituting the known values in it, we get
$
\Rightarrow x = \alpha \sin \left( {\dfrac{{2\pi }}{T}t + \phi } \right) \\
\Rightarrow 0 = \alpha \sin \left( {\dfrac{{2\pi }}{4} + \phi } \right) \\
 $
By simplification of the above step,
$\Rightarrow \phi = - \dfrac{\pi }{4}$
In order to find the amplitude of the particle travelling in the simple harmonic motion, use the same equation to find the value.
$\Rightarrow x = \alpha \sin \left( {\dfrac{{2\pi }}{T}t + \phi } \right)$
Substitute that $t = 4\,s$ $v = 4\,m{s^{ - 1}}$ and $\phi = - \dfrac{\pi }{4}$ in the above equation, we get
$\Rightarrow x = \alpha \sin \left( {\dfrac{{2\pi }}{{16}}4 - \dfrac{\pi }{4}} \right)$ …………………………..(1)
Differentiating the equation (1) with respect to the time taken,
$
\Rightarrow \dfrac{{dx}}{{dt}} = \alpha \sin \left( {\dfrac{\pi }{4}} \right) \\
  v = \dfrac{{dx}}{{dt}} = \alpha \times \dfrac{1}{{\sqrt 2 }} \\
 $
By performing the various arithmetic and the trigonometric operations, we get
$\Rightarrow \alpha = \dfrac{{32\sqrt 2 }}{\pi }$.
Hence the amplitude of the particle undergoing the simple harmonic motion is obtained as $\dfrac{{32\sqrt 2 }}{\pi }$ .

Thus the option (C) is correct.

Note: Understand the difference between the amplitude and the displacement. The displacement is the total distance covered by the particle in a straight line. If we differentiate it with respect to time, the velocity is obtained. The amplitude is the maximum height of the particle in the simple harmonic motion.