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 any particle in air when a tuning fork is vibrating, we can use the relationship between amplitude, frequency, and maximum velocity in simple harmonic motion.

### Step-by-step Solution:

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

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

3. **Calculate Maximum Velocity (v_max)**:
- The maximum velocity (v_max) of a particle in simple harmonic motion is given by:
\[ v_{max} = \omega \times A \]
- Substituting the values of ω and A:
\[ v_{max} = (500\pi) \times (0.01) \text{ m} \]
- Calculate:
\[ v_{max} = 5\pi \text{ m/s} \]

4. **Final Answer**:
- Therefore, the maximum velocity of any particle in air is:
\[ v_{max} = 5\pi \text{ m/s} \]