A sound wave of frequency `440 Hz` is passing through in air. An `O_(2)` molecule `( mass = 5.3 xx 10^(- 26) kg)` is set in oscillation with an amplitude of `10 ^(-6) m`. Its speed at the centre of its oscillation is

Options

A. `1.70 xx 10^(-5) m//s`
B. `17.0 xx 10^(-5) m//s`
C. `2.76 xx 10^(-3) m//s`
D. `2.77 xx 10^(-5) m//s`

Explanation

To find the speed of the oxygen molecule at the center of its oscillation, we can use the formula for the maximum speed in simple harmonic motion (SHM). The maximum speed (V_max) of an oscillating particle is given by:

\[ V_{\text{max}} = \omega \cdot A \]

where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude of oscillation.

1. **Calculate Angular Frequency (\( \omega \))**:
The angular frequency can be calculated using the formula:
\[ \omega = 2 \pi f \]
where \( f \) is the frequency of the sound wave. Given that \( f = 440 \, \text{Hz} \):
\[ \omega = 2 \pi \cdot 440 \, \text{Hz} \approx 2764.6 \, \text{rad/s} \]

2. **Use the Amplitude (\( A \))**:
The amplitude \( A \) is given as \( 10^{-6} \, \text{m} \).

3. **Calculate Maximum Speed (\( V_{\text{max}} \))**:
Now substitute \( \omega \) and \( A \) into the formula for maximum speed:
\[ V_{\text{max}} = \omega \cdot A = 2764.6 \, \text{rad/s} \cdot 10^{-6} \, \text{m} \]
\[ V_{\text{max}} \approx 2.7646 \times 10^{-3} \, \text{m/s} \]

4. **Final Result**:
Therefore, the speed of the oxygen molecule at the center of its oscillation is approximately:
\[ V_{\text{max}} \approx 2.76 \times 10^{-3} \, \text{m/s} \]

### Summary:
The speed of the oxygen molecule at the center of its oscillation is approximately \( 2.76 \times 10^{-3} \, \text{m/s} \).

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