Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

A particle is free to move on x-axis, in which of the following case, the particle will execute oscillation about $x = $ ?
a) $F = (x-1)$
b) $F = - (x-1)^{2}$
c) $F = - (x-1)^{3}$
d) $F = (x-1)^{3}$

seo-qna
Last updated date: 15th Jul 2024
Total views: 346.8k
Views today: 7.46k
Answer
VerifiedVerified
346.8k+ views
Hint: Let us consider a particle which is free to move on x-axis, then restoring force acts on the body i.e., $F =-kx^{n}$ . If n will be odd. Force should be along the positive x-axis for negative points on x-axis. Force should be along the negative x-axis for positive points on x-axis. Force should be zero for zero on x-axis. Then the particle will move to oscillate about a given point.

Complete step-by-step solution:
a) Given: $F = (x-1)$
When, $x = 1; F = 0$
When, $x > 1; F = +ve$
When, $x < 1; F = -ve$
This equation resembles with the equation of Simple Harmonic motion. Hence, the motion is Simple Harmonic.
b) Given: $F = - (x-1)^{2}$
When, $x = 1; F = 0$
When, $x > 1; F = -ve$
When, $x < 1; F = -ve$
In this case, motion is rectilinear motion. In rectilinear motion, particle move along a straight line.
c) Given: $F = - (x-1)^{3}$
When, $x = 1; F = 0$
When, $x > 1; F = -ve$
When, $x < 1; F = +ve$
In this case, motion is oscillatory about $x = 1$.
d) Given: $F = (x-1)^{3}$
When, $x = 1; F = 0$
When, $x > 1; F = +ve$
When, $x < 1; F = -ve$
So, this motion is not oscillatory.
Option (c) will be correct.

Note:Let us consider a particle which is free to move on x-axis, then restoring force acts on the body i.e., $F = -k x^{n}$ . If n will be even. Force should be along the negative x-axis for negative and positive points on the x-axis. Then the particle will not oscillate about a given point but will move rectilinearly.