A body is executing SHM with an amplitude of 0.1 m. Its velocity while passing through the mean position is `3ms^(-1)`. Its frequency in Hz is

Options

A. 15`pi`
B. `(15)/(pi)`
C. `30 pi`
D. `25 pi`

Explanation

To find the frequency of a body executing simple harmonic motion (SHM), we can use the relationship between the maximum velocity, amplitude, and angular frequency. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Amplitude (A) = 0.1 m
- Maximum velocity (v_max) = 3 m/s

### Step 2: Use the formula for maximum velocity in SHM
The maximum velocity (v_max) in SHM is given by the formula:
\[ v_{max} = \omega A \]
where:
- \( \omega \) is the angular frequency in radians per second,
- \( A \) is the amplitude.

### Step 3: Rearrange the formula to find angular frequency (ω)
From the formula, we can express angular frequency as:
\[ \omega = \frac{v_{max}}{A} \]

### Step 4: Substitute the known values
Now, substitute the known values into the equation:
\[ \omega = \frac{3 \, \text{m/s}}{0.1 \, \text{m}} \]
\[ \omega = 30 \, \text{rad/s} \]

### Step 5: Relate angular frequency to frequency (f)
The relationship between angular frequency (ω) and frequency (f) is given by:
\[ \omega = 2\pi f \]

### Step 6: Rearrange to find frequency (f)
Rearranging the formula gives:
\[ f = \frac{\omega}{2\pi} \]

### Step 7: Substitute the value of ω
Now substitute the value of ω:
\[ f = \frac{30 \, \text{rad/s}}{2\pi} \]
\[ f = \frac{30}{2\pi} \]
\[ f = \frac{15}{\pi} \]

### Step 8: Final result
Thus, the frequency of the body executing SHM is:
\[ f \approx 4.77 \, \text{Hz} \] (if you calculate \( \frac{15}{\pi} \) numerically)