Answer
64.8k+ views
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.
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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)