Question

A student says that he had applied a force F=-k√x on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he was 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

Answer:

b]x inc then k inc

Explanation:

f=-kx^1/2

i am using ~for proportionality

f`~kx^1/2

so,k~f/x^1/2

therefore, k~1/x^1/2

so k is inversely proportional to 1/x^1/2

so we can say b is ans

Answer 2

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

Explanation:

It is considered that a body is in simple harmonic motion only when,

$F=-k x$                         …(1)

here, force is denoted as F

Force constant is shown as k, and

body’s displacement from the mean position is shown as x.

It is shown:

$F=-k$

x                        …(2)

When the equations (1) and (2) are compared, it can be seen that k should be proportional to x to perform simple harmonic motion. So, k increases as x increases.