Question

A particle executing S.H.M. starts from the mean position. Its amplitude is $A$ and time period is $T$. At what displacement is its speed one-fourth of the maximum speed?

• A. $\frac{A\sqrt{15}}{4}$
• B. $\frac{A}{4}$
• C. $\frac{4A}{\sqrt{15}}$
• D. $\frac{A\sqrt{15}}{2}$

Hint:

You can use a normalized energy approach to solve this quickly: if velocity drops to $\frac{1}{4}$, the kinetic energy drops to $(\frac{1}{4})^2 = \frac{1}{16}$ of the total energy. Since total energy is conserved ($K + U = 1$), the potential energy must make up the remaining fraction: $U = 1 - \frac{1}{16} = \frac{15}{16}$. Since potential energy scales with $x^2$, the displacement must be $\sqrt{\frac{15}{16}}A = \frac{\sqrt{15}}{4}A$.

Answer 1

The Correct Option is A

Step 1: Recall the speed relation in SHM.
For a particle of amplitude $A$, the speed at displacement $x$ is $$v = \omega\sqrt{A^2 - x^2},$$ and the maximum speed (at the mean position, $x = 0$) is $v_{\max} = \omega A.$
Step 2: Write the given condition.
We want the displacement where $$v = \frac{1}{4}v_{\max}.$$
Step 3: Substitute both expressions.
$$\omega\sqrt{A^2 - x^2} = \frac{1}{4}\,\omega A.$$
Step 4: Cancel $\omega$ and square.
Dividing out $\omega$ and squaring both sides, $$A^2 - x^2 = \frac{A^2}{16}.$$
Step 5: Solve for $x^2$.
$$x^2 = A^2 - \frac{A^2}{16} = \frac{16A^2 - A^2}{16} = \frac{15A^2}{16}.$$
Step 6: Take the square root.
$$x = \frac{\sqrt{15}\,A}{4} = \frac{A\sqrt{15}}{4}.$$
\[ \boxed{x = \frac{A\sqrt{15}}{4}} \]