A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm / s . The frequency of its oscillation is

Options

A. 1 Hz
B. 3 Hz
C. 2 Hz
D. 4 Hz

Explanation

To find the frequency of a particle executing simple harmonic motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Amplitude (A) = 5 cm
- Maximum Speed (V_max) = 31.4 cm/s

2. **Use the Formula for Maximum Speed in SHM:**
The maximum speed (V_max) of a particle in SHM is given by the formula:
\[ V_{max} = \omega A \]
where \( \omega \) is the angular frequency.

3. **Rearrange the Formula to Solve for Angular Frequency (ω):**
We can rearrange the formula to find \( \omega \):
\[ \omega = \frac{V_{max}}{A} \]

4. **Substitute the Given Values:**
Substitute the values of \( V_{max} \) and \( A \):
\[ \omega = \frac{31.4 \, \text{cm/s}}{5 \, \text{cm}} = 6.28 \, \text{s}^{-1} \]

5. **Relate Angular Frequency to Frequency (f):**
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2\pi f \]

6. **Rearrange to Solve for Frequency (f):**
Rearranging gives:
\[ f = \frac{\omega}{2\pi} \]

7. **Substitute the Value of ω:**
Substitute the value of \( \omega \):
\[ f = \frac{6.28}{2\pi} \]

8. **Calculate the Frequency:**
Using \( \pi \approx 3.14 \):
\[ f = \frac{6.28}{2 \times 3.14} = \frac{6.28}{6.28} = 1 \, \text{Hz} \]

### Final Answer:
The frequency of the oscillation is \( 1 \, \text{Hz} \).