A particle of mass m is under the influence o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that,the force acting on the particle is $F = -kx + F_0$. We can rewrite this as $F = -k(x - \frac{F_0}{k})$. Let $y = x - \frac{F_0}{k}$,th

Vedclass · Accepted Answer

about $x = \frac{F_0}{k}$ with $\omega = \sqrt{\frac{k}{m}}$

Vedclass · Answer

(C) Given that,the force acting on the particle is $F = -kx + F_0$. We can rewrite this as $F = -k(x - \frac{F_0}{k})$. Let $y = x - \frac{F_0}{k}$,then the force becomes $F = -ky$. This is the standard form of the restoring force for Simple Harmonic Motion $(SHM)$,where the mean position is defined by $F = 0$. Setting $F = 0$,we get $-k(x - \frac{F_0}{k}) = 0$,which implies $x = \frac{F_0}{k}$. Thus,the particle oscillates about the mean position $x = \frac{F_0}{k}$. Comparing $F = -ky$ with the standard $SHM$ equation $F = -m\omega^2 y$,we get $m\omega^2 = k$. Therefore,the angular frequency is $\omega = \sqrt{\frac{k}{m}}$.

$A$ particle of mass $m$ is under the influence of a force $F = (-kx + F_0) \text{ N}$. The particle,when disturbed,will oscillate

Similar Questions

Which of the following equations does not represent a simple harmonic motion?

For a particle performing linear $SHM$,its average speed over $1$ oscillation is ($A =$ amplitude of $SHM$,$n =$ frequency of oscillation). (in $A \ n$)

The displacement of a particle is given at time $t$,by: $x = A \sin (-2 \omega t) + B \sin^2 \omega t$. Then,

The function of time representing a simple harmonic motion with a period of $\frac{\pi}{\omega}$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module