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?
(A) ${a^2}{T^2} + 4{\pi ^2}{v^2}$
(B) $\dfrac{{aT}}{x}$
(C) $aT + 2\pi \nu $
(D) $\dfrac{{aT}}{v}$

Answer 1

Hint: To answer this question, we have to consider the sinusoidal form of all the three quantities given. Then, substituting these sinusoidal forms in each option, we have to check which one is independent of time.

Complete step by step solution:
We know that for the simple harmonic motion of a particle, the variation of the displacement, the velocity and the acceleration is sinusoidal with time.
So, we can assume the displacement as
$\Rightarrow x = A\sin \left( {\omega t + \theta } \right)$ …………………...(i)
We know that $v = \dfrac{{dx}}{{dt}}$
Differentiating (i) with respect to time
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{{d\left[ {A\sin \left( {\omega t + \theta } \right)} \right]}}{{dt}}$
This gives
$\Rightarrow v = A\omega \cos \left( {\omega t + \varphi } \right)$ …………………….(ii)
Also, $a = \dfrac{{dv}}{{dt}}$
Differentiating (ii) with respect to time
$\Rightarrow \dfrac{{dv}}{{dt}} = \dfrac{{d\left[ {A\omega \cos \left( {\omega t + \theta } \right)} \right]}}{{dt}}$
$\Rightarrow a = - A{\omega ^2}\sin \left( {\omega t + \varphi } \right)$ …………………….(iii)
Considering the expression of option A
$\Rightarrow E = {a^2}{T^2} + 4{\pi ^2}{v^2}$
From (i) and (iii)
$\Rightarrow E = {\left( { - A{\omega ^2}\sin \left( {\omega t + \varphi } \right)} \right)^2}{T^2} + 4{\pi ^2}{\left( {A\omega \cos \left( {\omega t + \varphi } \right)} \right)^2}$ $E = {A^2}{\omega ^4}{T^2}{\sin ^2}\left( {\omega t + \varphi } \right) + 4{\pi ^2}{A^2}{\omega ^2}{\cos ^2}\left( {\omega t + \varphi } \right)$
Taking ${A^2}{\omega ^2}$ common, we get
$\Rightarrow E = {A^2}{\omega ^2}\left[ {{\omega ^2}{T^2}{{\sin }^2}\left( {\omega t + \varphi } \right) + 4{\pi ^2}{{\cos }^2}\left( {\omega t + \varphi } \right)} \right]$
Substituting $\omega = \dfrac{{2\pi }}{T}$
$\Rightarrow E = {A^2}{\omega ^2}\left[ {{{\left( {\dfrac{{2\pi }}{T}} \right)}^2}{T^2}{{\sin }^2}\left( {\omega t + \varphi } \right) + 4{\pi ^2}{{\cos }^2}\left( {\omega t + \varphi } \right)} \right]$
$\Rightarrow E = {A^2}{\omega ^2}\left[ {4{\pi ^2}{{\sin }^2}\left( {\omega t + \varphi } \right) + 4{\pi ^2}{{\cos }^2}\left( {\omega t + \varphi } \right)} \right]$
Taking $4{\pi ^2}$common, we get
$\Rightarrow E = 4{\pi ^2}{A^2}{\omega ^2}\left[ {{{\sin }^2}\left( {\omega t + \varphi } \right) + {{\cos }^2}\left( {\omega t + \varphi } \right)} \right]$
We know that${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\Rightarrow E = 4{\pi ^2}{A^2}{\omega ^2}$
As we can clearly see that this expression is independent of the time.
Hence, option A is correct.
Now, considering the expression of the option B
$\Rightarrow E = \dfrac{{aT}}{x}$
From (i) and (iii)
$\Rightarrow E = \dfrac{{\left[ { - A{\omega ^2}\sin \left( {\omega t + \varphi } \right)} \right]T}}{{A\sin \left( {\omega t + \varphi } \right)}}$
$\Rightarrow E = - {\omega ^2}T$
So, this expression is also independent of the time.
Hence, option B is also correct.
Now, considering the expression of option C
$\Rightarrow E = aT + 2\pi \nu $
We know that$\omega = 2\pi \nu $
$\Rightarrow E = aT + \omega $
From (iii)
$\Rightarrow E = - A{\omega ^2}T\sin \left( {\omega t + \varphi } \right) + \omega $
We can see that a constant term is being added to a time dependent term, which makes the whole expression dependent on time.
Hence, option C is incorrect.
Finally, considering the expression of option D
$\Rightarrow E = \dfrac{{aT}}{v}$
From (i) and (ii)
$\Rightarrow E = \dfrac{{ - A{\omega ^2}T\sin \left( {\omega t + \varphi } \right)}}{{A\omega \cos \left( {\omega t + \varphi } \right)}}$
$\Rightarrow E = \dfrac{{ - \omega T\sin \left( {\omega t + \varphi } \right)}}{{\cos \left( {\omega t + \varphi } \right)}}$
We know that $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $ . So we have
$\Rightarrow E = - \omega T\tan \left( {\omega t + \varphi } \right)$
So, this expression again comes out to be time dependent.
Hence, option D is also incorrect.
Option (A) and (B) are correct.

Note:
Instead of writing the complex sinusoidal equations for the three quantities discussed in the question, we can also directly use the relationship between these three quantities in SHM. The relationship is given as below
$\Rightarrow v = \omega \sqrt {{A^2} - {x^2}} $ and
$\Rightarrow a = - {\omega ^2}x$
Using this relationship, we can reduce our efforts and time.