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: Understanding the Question:
A particle undergoes simple harmonic motion with an amplitude $A$. We need to find the specific displacement ($x$) from the equilibrium position where the particle's instantaneous speed drops to exactly one-quarter of its maximum velocity ($v_{\text{max}}$).

Step 2: Key Formula or Approach:
1. The speed $v$ of a harmonic oscillator at any displacement position $x$ is given by the relation:
$$v = \omega \sqrt{A^2 - x^2}$$ 2. The maximum velocity ($v_{\text{max}}$) occurs as the particle passes through the mean equilibrium point ($x = 0$):
$$v_{\text{max}} = \omega A$$

Step 3: Detailed Explanation:
According to the problem statement, the instantaneous velocity satisfies the condition:
$$v = \frac{1}{4} v_{\text{max}}$$ Substitute the velocity formulas into this condition:
$$\omega \sqrt{A^2 - x^2} = \frac{1}{4} (\omega A)$$ We can cancel the angular frequency term $\omega$ from both sides:
$$\sqrt{A^2 - x^2} = \frac{A}{4}$$ Square both sides of the equation to eliminate the radical sign:
$$A^2 - x^2 = \left(\frac{A}{4}\right)^2 = \frac{A^2}{16}$$ Rearrange the terms to isolate the displacement variable $x^2$:
$$x^2 = A^2 - \frac{A^2}{16} = \frac{16A^2 - A^2}{16} = \frac{15A^2}{16}$$ Take the square root of both sides to find the displacement $x$:
$$x = \frac{\sqrt{15}A}{\sqrt{16}} = \frac{A\sqrt{15}}{4}$$ This expression matches the value given in option (A).

Step 4: Final Answer:
The displacement where the speed is one-fourth of the maximum speed is $\frac{A\sqrt{15}}{4}$, which corresponds to option (A).