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=10m/`s^(2)`

Options

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

Answer

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

### Step 1: Determine the frequency and time period
Given:
- Frequency (f) = 10 Hz

The time period (T) is the reciprocal of frequency:
\[ T = \frac{1}{f} = \frac{1}{10} = 0.1 \text{ seconds} \]

### Step 2: Relate time period to mass and spring constant
The time period of a mass-spring system in SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Squaring both sides, we have:
\[ T^2 = 4\pi^2 \frac{m}{k} \]
From this, we can express \(\frac{m}{k}\):
\[ \frac{m}{k} = \frac{T^2}{4\pi^2} \]

### Step 3: Substitute the time period into the equation
Substituting \(T = 0.1\) seconds:
\[ \frac{m}{k} = \frac{(0.1)^2}{4\pi^2} = \frac{0.01}{4\pi^2} \]

### Step 4: Find the angular frequency
The angular frequency (\(\omega\)) is given by:
\[ \omega = 2\pi f = 2\pi \times 10 = 20\pi \text{ rad/s} \]

### Step 5: Relate maximum speed to amplitude
The maximum speed (\(v_{max}\)) in SHM is given by:
\[ v_{max} = \omega A \]
where \(A\) is the amplitude.

### Step 6: Find the amplitude using the spring force
At the equilibrium position, the spring force balances the weight of the particle:
\[ kx_0 = mg \]
Thus, the amplitude \(A\) can be expressed as:
\[ A = \frac{mg}{k} \]

### Step 7: Substitute \(A\) into the maximum speed formula
Substituting \(A\) into the maximum speed equation:
\[ v_{max} = \omega \left(\frac{mg}{k}\right) \]

### Step 8: Substitute \(\frac{m}{k}\) into the equation
From step 3, we have \(\frac{m}{k} = \frac{0.01}{4\pi^2}\). Therefore:
\[ v_{max} = \omega \left(\frac{mg}{k}\right) = \omega \left(\frac{0.01}{4\pi^2} \cdot g\right) \]

### Step 9: Substitute values
Using \(g = 10 \, \text{m/s}^2\) and \(\omega = 20\pi\):
\[ v_{max} = 20\pi \left(\frac{0.01 \cdot 10}{4\pi^2}\right) \]
\[ = 20\pi \left(\frac{0.1}{4\pi^2}\right) \]
\[ = \frac{20 \cdot 0.1}{4\pi} = \frac{2}{\pi} \text{ m/s} \]

### Step 10: Final calculation
Calculating the numerical value:
\[ v_{max} \approx \frac{2}{3.14} \approx 0.636 \text{ m/s} \]

Thus, the maximum speed of the particle is approximately \(0.636 \text{ m/s}\).

### Final Answer
The maximum speed of the particle is \( \frac{1}{2\pi} \text{ m/s} \).