Question

A particle is suspended from a vertical spring which is executing S.H.M. of frequency $5\ \text{Hz}$. The spring is unstretched at the highest point of oscillation. Maximum speed of the particle is ($g = 10\ \text{m/s}^2$)

• A. $\frac{1}{\pi}\ \text{m/s}$
• B. $\frac{1}{4\pi}\ \text{m/s}$
• C. $\frac{1}{2\pi}\ \text{m/s}$
• D. $\pi\ \text{m/s}$

Hint:

For vertical SHM where the highest point is the natural length of the spring, the amplitude $A$ is exactly equal to the static extension of the spring under the particle's weight ($A = mg/k = g/\omega^2$).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A particle on a vertical spring executes SHM. We are given the frequency and the condition that at its highest point, the spring is at its natural, unstretched length. We need to find the particle's maximum speed.

Step 2: Key Formula or Approach:
At the highest point of oscillation, the spring is unstretched, meaning there is zero spring force. Thus, the only force providing the restoring acceleration is gravity. Therefore, the maximum acceleration $a_{\max} = A\omega^2$ must equal $g$.
The maximum speed is given by $v_{\max} = A\omega$.

Step 3: Detailed Explanation:
Equating the maximum acceleration to $g$:
$$A\omega^2 = g$$
$$A = \frac{g}{\omega^2}$$
The angular frequency $\omega$ is calculated from the given frequency $f = 5\ \text{Hz}$:
$$\omega = 2\pi f = 2\pi(5) = 10\pi\ \text{rad/s}$$
Substitute $\omega$ and $g = 10\ \text{m/s}^2$ into the amplitude equation:
$$A = \frac{10}{(10\pi)^2} = \frac{10}{100\pi^2} = \frac{1}{10\pi^2}\ \text{m}$$
Now, calculate the maximum speed $v_{\max}$:
$$v_{\max} = A\omega = \left(\frac{1}{10\pi^2}\right) \times (10\pi)$$
$$v_{\max} = \frac{1}{\pi}\ \text{m/s}$$

Step 4: Final Answer:
The maximum speed is $\frac{1}{\pi}\ \text{m/s}$, matching option (A).