Question

A particle executes S.H.M. of period $\frac{2\pi}{\sqrt{3}}$ second along a straight line $4\ \text{cm}$ long. The maximum velocity of the particle is

• A. $\sqrt{3}\ \text{cm/s}$
• B. $2\sqrt{3}\ \text{cm/s}$
• C. $4\sqrt{3}\ \text{cm/s}$
• D. $\frac{\sqrt{3}}{2}\ \text{cm/s}$

Hint:

A very common trap is to confuse the "length of the straight line path" directly with the amplitude. The total path length represents the entire range of motion, which spans from $-A$ to $+A$, making it $2A$. Always halve the path length to get your amplitude!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the time period of a simple harmonic oscillator and the total length of its linear path. We need to determine its maximum velocity.

Step 2: Key Formula or Approach:
The maximum velocity ($v_{\max}$) of a particle in 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.
The angular frequency $\omega$ is related to the time period $T$ by $\omega = \frac{2\pi}{T}$.
The total length of the straight-line path is exactly twice the amplitude ($2A$).

Step 3: Detailed Explanation:
First, find the amplitude $A$:
The path length is the distance between the two extreme positions, which is $2A = 4\ \text{cm}$.
$$A = \frac{4}{2} = 2\ \text{cm}$$
Next, calculate the angular frequency $\omega$ using the given time period $T = \frac{2\pi}{\sqrt{3}}\ \text{s}$:
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{\frac{2\pi}{\sqrt{3}}}$$
$$\omega = \sqrt{3}\ \text{rad/s}$$
Finally, calculate the maximum velocity:
$$v_{\max} = A \times \omega$$
$$v_{\max} = 2 \times \sqrt{3} = 2\sqrt{3}\ \text{cm/s}$$

Step 4: Final Answer:
The maximum velocity is $2\sqrt{3}\ \text{cm/s}$, matching option (B).