The frequency of a tuning fork with an amplitude A = 1 cm is 250 Hz. The maximum velocity of any particle in air is equal to

Options

A. 2.5 m/s
B. `5pi` m/s
C. `(3.30//pi)` m/sec
D. none of these

Answer

To find the maximum velocity of a particle in air due to the tuning fork, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Amplitude (A) = 1 cm = 0.01 m (since 1 cm = 0.01 m)
- Frequency (f) = 250 Hz

2. **Calculate Angular Frequency (ω)**:
- The angular frequency (ω) is related to the frequency (f) by the formula:
\[ \omega = 2\pi f \]
- Substitute the value of frequency:
\[ \omega = 2\pi \times 250 \text{ Hz} \]
- Calculate ω:
\[ \omega = 500\pi \text{ rad/s} \]

3. **Use the Formula for Maximum Velocity (V_max)**:
- The maximum velocity (V_max) of a particle in a wave can be calculated using the formula:
\[ V_{\text{max}} = A \cdot \omega \]
- Substitute the values of A and ω:
\[ V_{\text{max}} = 0.01 \text{ m} \cdot 500\pi \text{ rad/s} \]

4. **Calculate V_max**:
- Multiply the values:
\[ V_{\text{max}} = 0.01 \cdot 500\pi = 5\pi \text{ m/s} \]

5. **Final Result**:
- The maximum velocity of any particle in air is:
\[ V_{\text{max}} \approx 15.71 \text{ m/s} \quad (\text{using } \pi \approx 3.14) \]

### Summary:
The maximum velocity of any particle in air due to the tuning fork is \( 5\pi \) m/s.

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