Question

A particle of mass m kept at the origin is subjected to a force $\vec F = \left( {pt - qx} \right)\widehat i$ where t is the time elapsed and x is the coordinate of the position of the particle. Particle starts its motion at t=0 with zero initial velocity. If p and q are positive constants, then:
A. The acceleration of the particle will continuously keep on increasing with time
B. Particle will execute simple harmonic motion
C. The force on the particle will have no upper limit
D. The acceleration of particle will vary sinusoidally with time

Answer 1

Hint: In order to find the solution of the given question we will substitute the expression of force in the formula $F = ma$ then making the acceleration as our subject we will differentiate it with respect to time.

Formula used:
$F = ma$

Complete step-by-step answer:
Magnitude of the given force will be,
$F = \left( {pt - qx} \right)$
$ \Rightarrow F = pt - qx$
We know that force is defined as a push or pull on an object with mass that causes it to change its velocity. It is an external agent capable of changing the state of rest or motion of a particular body. It has a magnitude and a direction. The direction towards which the force is applied is referred to as the direction of the force.
Mathematically it is represented as, $F = ma$ where m is mass of the object and a is the acceleration.
$ \Rightarrow a = \dfrac{{pt - qx}}{m}$ --(1)
Now, differentiating the given expression with respect to time we get
$ \Rightarrow \dfrac{{da}}{{dt}} = \dfrac{{p - qv}}{m}$
$ \Rightarrow \dfrac{{{d^2}a}}{{d{t^2}}} = \dfrac{{ - q}}{m}\dfrac{{dv}}{{dt}}$
$ \Rightarrow \dfrac{{{d^2}a}}{{d{t^2}}} = \dfrac{{ - q}}{m}a$
Now, $\dfrac{{{d^2}a}}{{d{t^2}}} = - {\omega ^2}a$ --(2)
So, substituting the value of $\dfrac{{{d^2}a}}{{d{t^2}}} = - {\omega ^2}a$ we will get,
 $ \Rightarrow {\omega ^2} = \dfrac{q}{m}$
$ \Rightarrow \omega = \sqrt {\dfrac{q}{m}} $
Since, the particle will experience simple harmonic motion, so, time period is given by,
$T = \dfrac{{2\pi }}{\omega }$
substituting the value of $\omega $ we will get,
$T = 2\pi \sqrt {\dfrac{m}{q}} $
According to equation (1) we can say that the acceleration of the particle will continuously keep on increasing with time.
In equation (2) we see that the acceleration of the particle is equal to the negative of the displacement which means the particle will execute simple harmonic motion.
From the given expression of force, we can conclude that the force on the particle will have no upper limit.

So, the correct answers are “Option A, B and C”.

Note: We can find the solution of these types of questions by checking all the options one by one. Simple harmonic motion is defined as a type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position.