Question

Jerk is defined as the rate of change of acceleration of a particle. The velocity of a particle is $v = {v_0}\sin \omega t$ , where ${v_0}$and $\omega$are constant, Find jerk as a function of time.
a. ${v_0}{\omega ^2}\sin \omega t$
b. $- {v_0}{\omega ^2}\cos \omega t$
c. $- {v_0}{\omega ^2}\sin \omega t$
d. ${v_0}{\omega ^2}\cos \omega t$

Answer 1

The rate of change of acceleration is called jerk. This rate of change of acceleration can be calculated by differentiating the acceleration with respect to time. Since acceleration is the rate of change of velocity, it can be calculated by differentiating the velocity with respect to time. Therefore, to get the function for jerk the expression for velocity must be double differentiated.

Complete step by step answer:
It is given in the question that, the velocity of the particle is-
$v = {v_0}\sin \omega t$
Here, the velocity $v$is a function of time $t$. It is given that the terms ${v_0}$and $\omega$are constant. The term ${v_0}$ generally refers to the initial velocity of the object, whereas $\omega$ refers to the frequency or the angular speed of the particle. At an instant, $\omega$ multiplied with the instantaneous time $t$ gives the amount of angle covered by the object.

Therefore the acceleration of this particle can be determined by differentiating this function with respect to time,
$a = \dfrac{{dv}}{{dt}}$
$\Rightarrow a = \dfrac{d}{{dt}}\left( {{v_0}\sin \omega t} \right)$
The equation for derivative of $\sin x$ is,
$\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
Using this value in the above equation we get,
$a = {v_0}(\cos \omega t) \times \omega$
$\Rightarrow a = {v_0}\omega \cos \omega t$

Now, to calculate the jerk of the object, the expression of acceleration needs to be differentiated with respect to time.
Therefore,
$j = \dfrac{{da}}{{dt}}$or $j = \dfrac{{{d^2}v}}{{d{t^2}}}$
$\Rightarrow j = \dfrac{d}{{dt}}\left( {{v_0}\omega \cos \omega t} \right)$
The term ${v_0}\omega$ is constant. Therefore,
$j = {v_0}\omega \dfrac{d}{{dt}}\left( {\cos \omega t} \right)$
The equation for derivative of $\cos x$ is,
$\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$

Using this value in the above equation we get,
$j = {v_0}\omega \left( { - \sin \omega t} \right) \times \omega$
$\Rightarrow j = - {v_0}{\omega ^2}\sin \omega t$
The function defining the jerk of the particle is given by,
$j = - {v_0}{\omega ^2}\sin \omega t$

Hence, the correct answer is option (C).

Note: To calculate the jerk produced in a particle, the velocity must be double differentiated. Since velocity refers to the displacement of an object per unit time, the displacement of the object can also be used to calculate jerk. For this, the displacement of the particle must be differentiated three times. Thus, Jerk is the third derivative of displacement with respect to time.