Question

A particle starts simple harmonic motion from the mean position. Its amplitude is $a$ and total energy $E$. At an instant its kinetic energy is $\dfrac{3}{4}E$​. Its displacement at that instant is:
A) $\dfrac{a}{{\sqrt 2 }}$
B) $\dfrac{a}{2}$
C) $\sqrt 3 \dfrac{a}{2}$
D) Zero

Answer 1

Hint:
To find the values of displacement of the particle doing SHM at an instant asked in the question we can use the concept of kinetic energy of a particle doing SHM.



Formula used:
The particle's kinetic energy with mass $m$, equals:
$K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$

Complete step by step solution:
The particle's kinetic energy with mass $m$, equals:
$K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
While potential energy is the power that an object can store due to its position in relation to other things, internal tensions, electric charge, or other circumstances.
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$
As we know that the total energy is sum of the kinetic and potential energy, then we have:
 $E = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2}) + \dfrac{1}{2}m{\omega ^2}{x^2} \\$
$E = \dfrac{1}{2}m{\omega ^2}{a^2}\,\,\,\,....(i) \\$
Now, according to the question, the kinetic energy is $\dfrac{3}{4}E$​. Then we have:
$\dfrac{3}{4}E = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})\,\,\,\,....(ii)$
Compare the equation $(i)$from $(ii)$, then we obtain:
$1 - \dfrac{{{x^2}}}{{{a^2}}} = \dfrac{3}{4} \\$
 $\Rightarrow \dfrac{x}{a} = \sqrt {\dfrac{1}{4}} \\$
 $\Rightarrow \dfrac{x}{a} = \dfrac{1}{2} \\$
 $\Rightarrow x = \dfrac{a}{2} \\$
Thus, the correct option is:(B) $\dfrac{a}{2}$



Note:
It should be noted that we will have one negative value and one positive value when solving for the displacement. However, we have only thought about the positive root. This is due to the fact that the options only mentioned positive displacements.