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) given its amplitude and maximum speed, we can follow these steps:

### Step 1: Identify the given values
- Amplitude (A) = 5 cm
- Maximum speed (Vmax) = 31.4 cm/s

### Step 2: Use the formula for maximum speed in SHM
The maximum speed (Vmax) of a particle in SHM is given by the formula:
\[ V_{max} = A \cdot \omega \]
where:
- \( \omega \) is the angular frequency.

### Step 3: Rearrange the formula to find angular frequency
We can rearrange the formula to solve for \( \omega \):
\[ \omega = \frac{V_{max}}{A} \]

### Step 4: Substitute the known values
Now, substitute the known values into the equation:
\[ \omega = \frac{31.4 \, \text{cm/s}}{5 \, \text{cm}} \]
\[ \omega = 6.28 \, \text{s}^{-1} \]

### Step 5: Relate angular frequency to frequency
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2\pi f \]
We can rearrange this to find the frequency:
\[ f = \frac{\omega}{2\pi} \]

### Step 6: Substitute the value of \( \omega \)
Now, substitute the value of \( \omega \) we calculated:
\[ f = \frac{6.28 \, \text{s}^{-1}}{2\pi} \]
\[ f = \frac{6.28}{6.28} \]
\[ f = 1 \, \text{Hz} \]

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

---