Question

If E is the total energy of a particle executing SHM and ‘A’ is the amplitude of the vibratory motion, the E and ‘A’ are related as
$A. E \propto {A}^{2}$
$B. E \propto {1}/{{A}^{2}}$
$C. E \propto A$
$D. E \propto {1}/{A}$

Answer 1

Hint: To solve this question, first find the kinetic energy and the potential energy of the particle executing Simple Harmonic Motion (SHM). We know, total energy is the sum of the kinetic energy and potential energy. So, substitute the equations for kinetic energy and potential energy and add them up. Solve this equation and obtain the equation for total energy of the particle. From this obtained equation, find the relationship between E and A.

Complete step-by-step solution:
Kinetic energy of a particle executing SHM is given by,
$K=\dfrac {1}{2} m{\omega}^{2} \left({A}^{2}-{x}^{2} \right)$
Where,
m is the mass of the particle performing motion
A is the amplitude of the motion
x is the displacement of the particle from its mean position
${\omega}^{2}$ is the constant

Potential energy of a particle executing SHM is given by,
$V= \dfrac {1}{2}m{\omega}^{2}{x}^{2}$
Where,
m is the mass of the particle performing motion
x is the displacement of the particle from its mean position
${\omega}^{2}$ is the constant

Total energy of a particle executing SHM is given by,
$E= K+V$

Substituting values in above equation we get,
$E=\dfrac {1}{2} m{\omega}^{2} \left({A}^{2}-{x}^{2} \right)+ \dfrac {1}{2}m{\omega}^{2}{x}^{2}$
$\Rightarrow E= \dfrac {1}{2}m{\omega}^{2}{A}^{2}-\dfrac {1}{2}m{\omega}^{2}{x}^{2}+\dfrac {1}{2}m{\omega}^{2}{x}^{2}$
$\Rightarrow E= \dfrac {1}{2}m{\omega}^{2}{A}^{2}$
From the above equation, it is evident that,
$E \propto {A}^{2}$

So, the correct answer is option A i.e. $E \propto {A}^{2}$.

Note:
If the displacement of the particle from its mean position is zero, then the potential energy of the particle performing SHM becomes zero and is only dependent on the kinetic energy of the particle. If the displacement of the particle from its mean position is equal to the amplitude of the motion of the particle, then the kinetic energy of the particle performing SHM becomes zero and is only dependent on the potential energy of the particle.