Question

A particle of mass $0.1kg$ executes SHM under a force $F=(-10x)N$. Speed of particle at mean position is $6m/s.$Then amplitude of oscillations is

• A. $0.6m$
• B. $0.2m$
• C. $0.4m$
• D. $0.1m$

Answer 1

The correct option is A

$0.6m$

Step 1. Given Data,

The force causing the simple harmonic motion is $F=(-10x)$N

The mass of the particles is $m=0.1kg$

Step 2. Formula used:
$v_{max}=A\omega$
Where $v_{max}$is the maximum velocity of a particle oscillating in SHM,
$A$is the amplitude of the particle.
$\omega$ is the angular velocity of the particle.
Step 3. Calculating the amplitude of oscillation,
From Newton’s second law of motion,
We know that-
$F=ma$
$a$ is the acceleration of the body.
On substituting the given values,
$-10x=0.1a$
$a=-100x$
For a simple harmonic motion,
$a=-{\omega }^{2}x$
Comparing the value of acceleration in both cases,
$a=-{\omega }^{2}x=-100x$
We get,
$\omega =10rad/\mathrm{s}$
We know that the maximum velocity of the particle is given by,
$\left|v_{max}\right|=\left|-A\omega \right|$
From the question, $v_{max}=6m/s$
Thus,
$A={\displaystyle \frac{v_{max}}{\omega }}$
$A={\displaystyle \frac{6}{10}}=0.6$
Therefore, the amplitude of the oscillation of the particle is $0.6m$.

Hence option A is the correct answer.