Question

A student says that he had applied a force $F=-k\sqrt{x}$ on a particle and the particle moves in simple harmonic motion. He refuses to tell whether $k$ is a constant or not. Assume that he has worked only with positive $x$ and no other force acted on the particle.

• A. As $x$ increases $k$ increases
• B. As $x$ increases $k$ decreases
• C. As $x$ increases $k$ remains constant
• D. The motion cannot be simple harmonic

Answer 1

The correct option is A As $x$ increases $k$ increases

Given that the motion of the particle is Simple Harmonic Motion.

$F=-m{\omega }^{2}x$

The force applied by the student is given by,

$F=-\frac{-k}{x}$

If the motion is S.H.M., then both will be equal.

So,

$\frac{-k}{-x}=m{\omega }^{2}x$

$k=m{\omega }^2{x}^2$

This means, if $x$ increases, then $k$ also increases.

So the correct answer is option a.