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

Answer

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

3. **Relate Angular Frequency to Linear Frequency:**
The angular frequency (\( \omega \)) can be related to the linear frequency (N) by the equation:
\[ \omega = 2 \pi N \]
Therefore, we can rewrite the maximum speed formula as:
\[ V_{max} = A \cdot (2 \pi N) \]

4. **Rearranging the Formula to Solve for Frequency (N):**
We can rearrange the equation to solve for the frequency (N):
\[ N = \frac{V_{max}}{2 \pi A} \]

5. **Substituting the Known Values:**
Now, we substitute the known values into the equation:
\[ N = \frac{31.4 \, \text{cm/s}}{2 \cdot \pi \cdot 5 \, \text{cm}} \]

6. **Calculating the Frequency:**
First, calculate the denominator:
\[ 2 \cdot \pi \cdot 5 \approx 2 \cdot 3.14 \cdot 5 = 31.4 \]
Now substitute this back into the frequency equation:
\[ N = \frac{31.4}{31.4} = 1 \, \text{Hz} \]

7. **Final Answer:**
The frequency of the oscillation is:
\[ N = 1 \, \text{Hz} \]