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The amplitude of simple harmonic motion represented by the displacement equation $y\left( {cm} \right) = 4(sin5\pi t + \sqrt 2 \cos 5\pi t)$ is:
(A) $4cm$
(B) $4\sqrt 2 cm$
(C) $4\sqrt 3 cm$
(D) $4\sqrt {2 + 1} cm$

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Last updated date: 17th Apr 2024
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Answer
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Hint The amplitude of any motion is the maximum displacement of the body from its mean position. Hence we can say that the maximum value of the displacement in its equation of motion is the amplitude of that motion.

Formula used
For $f(x) = a\sin (x) + b\cos (x)$ the maximum value of $f(x)$ is $\sqrt {{a^2} + {b^2}} $
$\dfrac{{{d^2}x}}{{d{t^2}}} + {\omega ^2}x = 0$
Where $x$represents the displacement from mean position, $\omega $is the angular frequency of that simple harmonic motion, and $\dfrac{{{d^2}x}}{{d{t^2}}}$ represents the double derivative of displacement with respect to time.

Complete Step-by-step answer
A simple harmonic motion is a motion in which the restoring force is directly proportional to the displacement from the mean position and whose equation can be represented in the general equation of simple harmonic motion which is
$\dfrac{{{d^2}x}}{{d{t^2}}} + {\omega ^2}x = 0$
Where $x$represents the displacement from mean position, $\omega $is the angular frequency of that simple harmonic motion, and $\dfrac{{{d^2}x}}{{d{t^2}}}$ represents the double derivative of displacement with respect to time.
It’s given that equation of motion is,
$y\left( x \right) = 4(sin5\pi t + \sqrt 2 \cos 5\pi t)$
We know that,
For $f(x) = a\sin (x) + b\cos (x)$ the maximum value of $f(x)$ is $\sqrt {{a^2} + {b^2}} $
Hence
$ \Rightarrow y\left( x \right) = 4sin5\pi t + 4\sqrt 2 \cos 5\pi t$
On comparing both the equations $a = 4$ and $b = 4\sqrt 2 $
Hence the maximum value of $ \Rightarrow y\left( x \right) = 4sin5\pi t + 4\sqrt 2 \cos 5\pi t$is
$ \Rightarrow \sqrt {{4^2} + {{(4\sqrt 2 )}^2}} $
$ \Rightarrow \sqrt {{4^2} + 2{{(4)}^2}} $
$ \Rightarrow \sqrt {3{{(4)}^2}} $
$ \Rightarrow 4\sqrt 3 $
Therefore from the theory and the known formula we can say that the amplitude of the equation is $4\sqrt 3 cm$

Hence the correct answer to the above question is (C) $4\sqrt 3 cm$

Note
We used the known equation $f(x) = a\sin (x) + b\cos (x)$ the maximum value of $f(x)$ is $\sqrt {{a^2} + {b^2}} $ but it has proof and to understand it one must have basic knowledge of the application of derivatives.
In this first, we double differentiate $f(x)$ and find the point of local maximum and then we substitute it in $f(x)$ to find the local maxima which is the maximum value of our function.