Question

The maximum velocity of a particle performing S.H.M. is 'V'. If the periodic time is made $\left(\frac{1}{3}\right)^{rd}$ and the amplitude is doubled, then the new maximum velocity of the particle will be

• A. $\frac{V}{6}$
• B. $\frac{3V}{2}$
• C. $3V$
• D. $6V$

Hint:

Logic Tip: Use direct proportionality! $V_{max} \propto \frac{A}{T}$. If you multiply the numerator $A$ by 2 and multiply the denominator $T$ by $\frac{1}{3}$, the overall multiplier is $\frac{2}{1/3} = 6$. So, the new velocity is $6V$.

Answer 1

The Correct Option is D

Concept:
The maximum velocity ($V_{max}$) of a particle undergoing Simple Harmonic Motion (S.H.M.) occurs at the mean position and is given by the formula: $$V_{max} = A\omega$$ where $A$ is the amplitude and $\omega$ is the angular frequency. Angular frequency is related to the periodic time ($T$) by $\omega = \frac{2\pi}{T}$. Therefore: $$V_{max} = \frac{2\pi A}{T}$$
Step 1: Set up the initial condition.
Let the initial amplitude be $A$ and the initial time period be $T$. The initial maximum velocity is: $$V = A \left(\frac{2\pi}{T}\right)$$
Step 2: Define the new parameters.
The periodic time is made $\frac{1}{3}$ of its original value: $$T' = \frac{T}{3}$$ The amplitude is doubled: $$A' = 2A$$
Step 3: Calculate the new maximum velocity.
Substitute the new parameters into the maximum velocity formula: $$V' = \frac{2\pi A'}{T'}$$ $$V' = \frac{2\pi (2A)}{\left(\frac{T}{3}\right)}$$ Bring the 3 from the denominator up to the numerator: $$V' = 3 \times 2 \times \frac{2\pi A}{T}$$ $$V' = 6 \left(\frac{2\pi A}{T}\right)$$ Substitute the initial velocity $V$: $$V' = 6V$$