Question

A particle is executing simple harmonic motion. If the force acting on the particle at a position is 86.6% of the maximum force on it, then the ratio of its velocity at that point and its maximum velocity is

• A. $1:\sqrt{3}$
• B. $1:2$
• C. $\sqrt{3}:2$
• D. $1:3$

Hint:

Common SHM values: At $x = \frac{\sqrt{3}}{2}A$, $v = \frac{1}{2}v_{max}$. At $x = \frac{1}{2}A$, $v = \frac{\sqrt{3}}{2}v_{max}$. Energy partitions similarly.

Answer 1

The Correct Option is B

Step 1: Relate Force to Position:
Restoring force in SHM is $F = -kx$. Maximum force is $F_{max} = kA$. Given $F = 86.6% \text{ of } F_{max} = 0.866 F_{max}$. \[ kx = 0.866 kA \] \[ x = 0.866 A = \frac{\sqrt{3}}{2} A \] (Since $\frac{\sqrt{3}}{2} \approx 0.866$).
Step 2: Relate Velocity to Position:
Velocity at position $x$ is $v = \omega \sqrt{A^2 - x^2}$. Maximum velocity is $v_{max} = \omega A$.
Step 3: Calculate Ratio:
\[ \frac{v}{v_{max}} = \frac{\omega \sqrt{A^2 - x^2}}{\omega A} = \sqrt{1 - \left(\frac{x}{A}\right)^2} \] Substitute $\frac{x}{A} = \frac{\sqrt{3}}{2}$: \[ \frac{v}{v_{max}} = \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \] Ratio is $1:2$.