Frequency of a particle executing SHM is 10 Hz. The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is `(g=10(m)/(s))`

Options

A. `2pi(m)/(s)`
B. `pi(m)/(s)`
C. `(1)/(pi)(m)/(s)`
D. `(1)/(2pi)/(m)/(s)`

Explanation

To find the maximum speed of a particle executing simple harmonic motion (SHM) suspended from a vertical spring, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Frequency \( f = 10 \, \text{Hz} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Calculate Angular Frequency (\( \omega \))**:
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \times 10 = 20\pi \, \text{rad/s} \]

3. **Set Up the Equilibrium Condition**:
At the equilibrium position, the weight of the particle is balanced by the spring force:
\[ mg = kx \]
Here, \( x \) is the amplitude \( a \) of the motion. Thus, we can write:
\[ mg = ka \]

4. **Express Spring Constant (\( k \))**:
The angular frequency \( \omega \) is also related to the spring constant \( k \) and mass \( m \) by:
\[ \omega^2 = \frac{k}{m} \]
Rearranging gives:
\[ k = m\omega^2 \]

5. **Substitute for \( k \) in the Equilibrium Condition**:
Substitute \( k \) into the equilibrium condition:
\[ mg = m\omega^2 a \]
Cancel \( m \) (assuming \( m \neq 0 \)):
\[ g = \omega^2 a \]

6. **Solve for Amplitude (\( a \))**:
Rearranging gives:
\[ a = \frac{g}{\omega^2} \]
Substitute \( g = 10 \, \text{m/s}^2 \) and \( \omega = 20\pi \):
\[ a = \frac{10}{(20\pi)^2} = \frac{10}{400\pi^2} = \frac{1}{40\pi^2} \, \text{m} \]

7. **Calculate Maximum Speed (\( V_{\text{max}} \))**:
The maximum speed in SHM is given by:
\[ V_{\text{max}} = a \cdot \omega \]
Substitute the values of \( a \) and \( \omega \):
\[ V_{\text{max}} = \left(\frac{1}{40\pi^2}\right) \cdot (20\pi) = \frac{20\pi}{40\pi^2} = \frac{1}{2\pi} \, \text{m/s} \]

### Final Answer:
The maximum speed of the particle is:
\[ V_{\text{max}} = \frac{1}{2\pi} \, \text{m/s} \]