Answer
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Hint:- A simple harmonic motion is a periodic motion in which there is a restoring force which is acting towards the equilibrium position of the body also the restoring force is proportional to the displacement of the body executing simple harmonic motion.
Formula used: The formula of acceleration of the body executing simple harmonic motion is given by,
${a_{\max .}} = A{\omega ^2}$
Where ${a_{\max .}}$ is the maximum acceleration of the body A is the amplitude of the body and $\omega $ is the angular frequency of the body.
Complete step-by-step solution
It is given in the problem that an object executing a simple harmonic motion has an amplitude of 0.01 m and the frequency of the simple harmonic motion is 40 Hz and we need to find the value of the maximum acceleration of the body executing simple harmonic motion.
As the angular frequency of the body is given by,
$\omega = 2\pi f$
Where $\omega $ is the angular acceleration and f is the frequency of the body is 40 Hz.
So the angular frequency is given by,
$ \Rightarrow \omega = 2\pi f$
$ \Rightarrow \omega = 2\pi \cdot \left( {40} \right)$
$ \Rightarrow \omega = 80\pi \dfrac{1}{s}$………eq. (1)
As the formula of maximum acceleration is given by,
${a_{\max .}} = A{\omega ^2}$
Where ${a_{\max .}}$is the maximum acceleration of the body, A is the amplitude of the body and $\omega $ is the angular frequency of the body.
Replacing the value of the amplitude A and the angular frequency of the body executing simple harmonic motion we can get the value of maximum acceleration of the body.
$ \Rightarrow {a_{\max .}} = A{\omega ^2}$
Replace the value of amplitude $A = 0 \cdot 01m$ and $\omega = 80\pi $ we get,
$ \Rightarrow {a_{\max .}} = \left( {0 \cdot 01} \right){\left( {80\pi } \right)^2}$
$ \Rightarrow {a_{\max .}} = 630\dfrac{m}{{{s^2}}}$.
The maximum acceleration is equal to${a_{\max .}} = 630\dfrac{m}{{{s^2}}}$. The correct answer for this problem is option A.
Note:- The maximum acceleration of the body executing simple harmonic motion is independent of the mass of the body also it is independent of the displacement of the body which is executing simple harmonic motion.
Formula used: The formula of acceleration of the body executing simple harmonic motion is given by,
${a_{\max .}} = A{\omega ^2}$
Where ${a_{\max .}}$ is the maximum acceleration of the body A is the amplitude of the body and $\omega $ is the angular frequency of the body.
Complete step-by-step solution
It is given in the problem that an object executing a simple harmonic motion has an amplitude of 0.01 m and the frequency of the simple harmonic motion is 40 Hz and we need to find the value of the maximum acceleration of the body executing simple harmonic motion.
As the angular frequency of the body is given by,
$\omega = 2\pi f$
Where $\omega $ is the angular acceleration and f is the frequency of the body is 40 Hz.
So the angular frequency is given by,
$ \Rightarrow \omega = 2\pi f$
$ \Rightarrow \omega = 2\pi \cdot \left( {40} \right)$
$ \Rightarrow \omega = 80\pi \dfrac{1}{s}$………eq. (1)
As the formula of maximum acceleration is given by,
${a_{\max .}} = A{\omega ^2}$
Where ${a_{\max .}}$is the maximum acceleration of the body, A is the amplitude of the body and $\omega $ is the angular frequency of the body.
Replacing the value of the amplitude A and the angular frequency of the body executing simple harmonic motion we can get the value of maximum acceleration of the body.
$ \Rightarrow {a_{\max .}} = A{\omega ^2}$
Replace the value of amplitude $A = 0 \cdot 01m$ and $\omega = 80\pi $ we get,
$ \Rightarrow {a_{\max .}} = \left( {0 \cdot 01} \right){\left( {80\pi } \right)^2}$
$ \Rightarrow {a_{\max .}} = 630\dfrac{m}{{{s^2}}}$.
The maximum acceleration is equal to${a_{\max .}} = 630\dfrac{m}{{{s^2}}}$. The correct answer for this problem is option A.
Note:- The maximum acceleration of the body executing simple harmonic motion is independent of the mass of the body also it is independent of the displacement of the body which is executing simple harmonic motion.
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