A sound wave of frequency 440 Hz is passing through air. An `O_2` molecule (mass`=5.3xx10^(-26)Kg)` is set in oscillation with an amplitude of `10^-6m` speed at the centre of its oscillation is :

Options

A. `1.70xx10^(-6) m//s`
B. `17.0xx10^(-6) m//s`
C. `2.76 xx10^(-3) m//s`
D. `2.77xx10^(-6) m//s`

Explanation

To find the speed of an oxygen molecule oscillating in a sound wave at the center of its oscillation, we can use the formula for the maximum speed in simple harmonic motion (SHM):

### Step-by-Step Solution:

1. **Identify the given values:**
- Frequency of the sound wave, \( f = 440 \, \text{Hz} \)
- Amplitude of oscillation, \( A = 10^{-6} \, \text{m} \)

2. **Calculate the angular frequency (\( \omega \)):**
The angular frequency is given by the formula:
\[ \omega = 2 \pi f \]
Substituting the value of frequency:
\[ \omega = 2 \pi \times 440 \, \text{Hz} \]

3. **Calculate \( \omega \):**
\[ \omega \approx 2 \times 3.14 \times 440 \approx 2764.6 \, \text{rad/s} \]

4. **Calculate the maximum speed (\( v_{\text{max}} \)):**
The maximum speed in SHM is given by:
\[ v_{\text{max}} = \omega A \]
Substituting the values of \( \omega \) and \( A \):
\[ v_{\text{max}} = 2764.6 \times 10^{-6} \, \text{m/s} \]

5. **Calculate \( v_{\text{max}} \):**
\[ v_{\text{max}} \approx 2.7646 \times 10^{-3} \, \text{m/s} \]

6. **Final Answer:**
Rounding to three significant figures, the speed at the center of its oscillation is:
\[ v_{\text{max}} \approx 2.76 \times 10^{-3} \, \text{m/s} \]

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

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