Question

A particle is performing S.H.M. with maximum velocity $v$. If the amplitude is tripled and the periodic time is doubled, then the new maximum velocity will be

• A. $1.5\ v$
• B. $3\ v$
• C. $2\ v$
• D. $v$

Hint:

Using a direct ratio method saves setup time: since $v \propto \frac{A}{T}$, the scaling multiplier is simply $\frac{\text{Amplitude factor}}{\text{Time factor}}$. Here, plugging in the factors gives $\frac{3}{2} = 1.5$ immediately.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem tracks a particle performing Simple Harmonic Motion (S.H.M.) with an initial peak velocity $v$. We need to compute the new maximum velocity when its structural amplitude parameter is tripled and its oscillation period is doubled.

Step 2: Key Formula or Approach:
The maximum velocity ($v_{\max}$) achieved by an object in S.H.M. occurs at the mean position and is given by: $$v_{\max} = \omega A$$ Where $A$ is the amplitude and $\omega$ is the angular frequency. Since $\omega = \frac{2\pi}{T}$, we can rewrite the peak velocity expression in terms of the time period $T$: $$v_{\max} = \frac{2\pi A}{T} \implies v_{\max} \propto \frac{A}{T}$$

Step 3: Detailed Explanation:
Let the initial state parameters be amplitude $A_1 = A$ and period $T_1 = T$, giving an initial peak velocity: $$v = \frac{2\pi A}{T}$$ According to the problem statement, the parameters undergo the following modifications: New amplitude: $A_2 = 3A$ New time period: $T_2 = 2T$
Set up the formula for the new maximum velocity $v'$ using these updated values: $$v' = \frac{2\pi A_2}{T_2} = \frac{2\pi (3A)}{2T}$$ Factor out the numeric coefficients to see the proportional change clearly: $$v' = \left(\frac{3}{2}\right) \cdot \frac{2\pi A}{T}$$ Substitute the initial velocity expression ($v = \frac{2\pi A}{T}$) back into the equation: $$v' = \frac{3}{2}v = 1.5\ v$$

Step 4: Final Answer:
The new maximum velocity is $1.5\ v$, matching option (A).