A particle moving about its equilibrium position with equation $y = - ax - bt$. Interpret the condition.
A. It will always perform the SHM
B. It can never perform the SHM
C. It can perform SHM only when $t \geqslant \dfrac{{bx}}{a}$
D. It can perform SHM only when $t \leqslant \dfrac{{bx}}{a}$

Hint: We need to convert the given equation into a second order differential equation. Then we need to compare it with the standard equation for simple harmonic motion. The particle will execute SHM only if the two equations have the same form.

Complete answer:
We are given a particle which is moving about its equilibrium position. Its equation of motion is given as
$y = - ax - bt$
Here a and b are constants while x, y and t are variables.
We need to check if this particle is executing a simple harmonic motion or not. For the case of an object executing a simple harmonic motion, we have studied that such type of motion is described by the following type of equation.
$\dfrac{{{d^2}y}}{{d{x^2}}} + {\omega ^2}x = 0$
Here $\omega $ represents the angular frequency of the oscillations of the given particle. In order to check if the given executes this type of motion or not, we need to take the derivative of the given equation with respect to variable x. Doing so, we get
$\dfrac{{dy}}{{dx}} = - a - b\dfrac{{dt}}{{dx}}$
Now we will again take the derivative of this equation with respect to time. We get
$\dfrac{{{d^2}y}}{{d{x^2}}} = - b\dfrac{{{d^2}t}}{{d{x^2}}}$
We can write it as
$\dfrac{{{d^2}y}}{{d{x^2}}} + b\dfrac{{{d^2}t}}{{d{x^2}}} = 0$
When we compare this equation with the standard equation for simple harmonic motion, we notice that they are not identical. This means that the given particle does not execute any simple harmonic motion.

So, the correct answer is “Option B”.

Note:
If we put $t = \dfrac{{bx}}{a}$ in the equation for the given particle, we get $\dfrac{{{d^2}y}}{{d{x^2}}} = 0$ which is still not the equation for SHM. Therefore, the last two options are also wrong. The existence of a term containing time t in our equation does not allow the particle to execute the SHM.