The pressure wave, `P = 0.01 sin[1000t-3x]Nm^(-2)`, corresponds to the sound produced by a vibrating blabe on a day when a atmospheric temperature si `0^(@)C`. On some other day when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be `336 ms^(-1)`. Approximate value of T is :

Options

A. `4^(@)C`
B. `11^(@)C`
C. `12^(@)C`
D. `15^(@)C`

Answer

To solve the problem, we need to find the approximate temperature \( T \) when the speed of sound is \( 336 \, \text{m/s} \). We start with the given pressure wave equation and the relationship between the speed of sound and temperature.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The pressure wave is given by \( P = 0.01 \sin(1000t - 3x) \, \text{N/m}^2 \).
- The speed of sound at \( 0^\circ C \) is calculated from the wave parameters.
- Speed of sound at temperature \( T \) is \( 336 \, \text{m/s} \).

2. **Extract Parameters from the Wave Equation:**
- The angular frequency \( \omega = 1000 \, \text{rad/s} \).
- The wave number \( k = 3 \, \text{rad/m} \).

3. **Calculate the Speed of Sound at \( 0^\circ C \):**
- The speed of sound \( V \) is given by the formula:
\[ V = \frac{\omega}{k} \]
- Substituting the values:
\[ V = \frac{1000}{3} \approx 333.33 \, \text{m/s} \]

4. **Relate Speed of Sound to Temperature:**
- The speed of sound in air is related to the absolute temperature \( T \) by the formula:
\[ V \propto \sqrt{T} \]
- Thus, we can write:
\[ \frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} \]
- Where:
- \( V_1 = 333.33 \, \text{m/s} \) (speed at \( 0^\circ C \))
- \( V_2 = 336 \, \text{m/s} \)
- \( T_1 = 273 \, \text{K} \) (temperature at \( 0^\circ C \))
- \( T_2 \) is the temperature we need to find.

5. **Set Up the Equation:**
- Plugging in the values:
\[ \frac{333.33}{336} = \sqrt{\frac{273}{T_2}} \]

6. **Square Both Sides:**
- Squaring both sides gives:
\[ \left(\frac{333.33}{336}\right)^2 = \frac{273}{T_2} \]

7. **Calculate the Left Side:**
- Calculating \( \frac{333.33}{336} \):
\[ \frac{333.33}{336} \approx 0.992 \]
- Squaring this value:
\[ (0.992)^2 \approx 0.984 \]

8. **Solve for \( T_2 \):**
- Rearranging the equation:
\[ T_2 = \frac{273}{0.984} \]
- Calculating \( T_2 \):
\[ T_2 \approx 277.4 \, \text{K} \]

9. **Convert to Celsius:**
- To convert Kelvin to Celsius:
\[ T = T_2 - 273 \approx 277.4 - 273 \approx 4.4^\circ C \]

### Final Answer:
The approximate value of \( T \) is \( 4^\circ C \).