Question

A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is $\frac{3E}{4}$ then its displacement 'y' is given by :

• A. $y = \frac{a}{\sqrt{2}}$
• B. $y = \frac{a}{2}$
• C. $y = \frac{a\sqrt{3}}{2}$
• D. $y = a$

Hint:

A quick way to solve SHM energy problems is to use the energy ratios. Total energy is proportional to $a^2$. Potential energy is proportional to $y^2$. The relationship $PE/E = y^2/a^2$ is very useful.

Answer 1

The Correct Option is B

The total mechanical energy (E) of a particle in SHM is constant and is given by:
$E = \frac{1}{2} k a^2$, where k is the force constant and a is the amplitude.
The kinetic energy (KE) at a displacement y from the mean position is:
$KE = \frac{1}{2} k (a^2 - y^2)$
The potential energy (PE) at displacement y is:
$PE = \frac{1}{2} k y^2$
We are given that $KE = \frac{3E}{4}$.
We also know that Total Energy $E = KE + PE$.
Therefore, $PE = E - KE = E - \frac{3E}{4} = \frac{E}{4}$.
Now, we substitute the expressions for PE and E:
$\frac{1}{2} k y^2 = \frac{1}{4} \left( \frac{1}{2} k a^2 \right)$
Cancel the common term $\frac{1}{2}k$ from both sides:
$y^2 = \frac{1}{4} a^2$
Taking the square root of both sides:
$y = \pm \frac{a}{2}$
The displacement is given by $y = a/2$.