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# The phase difference between the displacement and acceleration of the particle executing SHM in radianA) $\dfrac{\pi }{4}$ B) $\dfrac{\pi }{2}$C) $\pi$ D) $2\pi$

Last updated date: 20th Jun 2024
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Hint: Acceleration of a particle executing simple harmonic motion can be given to be equal to the second-order derivative of the displacement of the particle. We know the displacement of a particle executing SHM is given in terms of its amplitude and sine of its phase.

We know that the equation of the SHM of a particle is given by:
$x = A\sin \omega \,t$
Here x is the displacement of the particle; A is the amplitude, $\omega$ is the angular frequency and t is the time.
After differentiating the above equation w.r.t time, we will get the velocity. It can be shown as:
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {A\sin \omega \,t} \right)$
$\Rightarrow v = A\,\omega \cos \omega \,t$ …… (I)
Here $\dfrac{{dx}}{{dt}} = v$ , it means that the differentiation of x with respect to t is velocity v.
Now we will differentiate the equation (I) again, and we get,
$\Rightarrow \dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}\left( {A\,\omega \cos \omega \,t} \right)$
$\Rightarrow a = - A\,{\omega ^2}\sin \omega \,t$
Here $\dfrac{{dv}}{{dt}} = a$, it means that the differentiation of v with respect to t is acceleration a.
Now, as we can notice that the displacement x has $\sin \omega \,t$ and the velocity v has $\cos \omega \,t$, we can say that the phase difference between x and v is 90 degrees $\dfrac{\pi }{2}$.
Also, we can notice that the velocity v has $\cos \omega \,t$ and the acceleration a has $\sin \omega \,t$ , and hence we can say that the phase difference between v and a is 90 degrees that is $\dfrac{\pi }{2}$.
From the above two statements, we can say that the phase difference between displacement x and acceleration a is,
$\Rightarrow \phi = \dfrac{\pi }{2} + \dfrac{\pi }{2}$
$\Rightarrow \phi = \pi$
Here $\phi$ is the phase difference, which is equal to $\pi$.

Therefore, the correct option is (C).

Note: We must be careful while comparing the equation given in the question to that of the general equation of SHM. Sometimes, instead of the value of angular frequency, the value of time may be provided; in that case, we need to compare the coefficients of the angular frequency and not time.