The equation of a sound wave at `0^(@)C` is given as `y=Asin(1000t-3x)` the speed at some other temperature T is given `336m//s` the value of T is

Options

A. `4.4^(@)C`
B. `11^(@)C`
C. `12^(@)C`
D. `7^(@)C`

Answer

To solve the problem, we need to find the temperature \( T \) at which the speed of sound is \( 336 \, \text{m/s} \), given the wave equation at \( 0^\circ C \) (or \( 273 \, \text{K} \)) is \( y = A \sin(1000t - 3x) \).

### Step-by-Step Solution:

1. **Identify the wave equation parameters**:
The wave equation is given as:
\[ y = A \sin(1000t - 3x) \]
From this, we can identify:
- Angular frequency \( \omega = 1000 \, \text{rad/s} \)
- Wave number \( k = 3 \, \text{rad/m} \)

2. **Calculate the speed of sound at \( 0^\circ C \)**:
The speed of sound \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
Substituting the values:
\[ v = \frac{1000}{3} \approx 333.33 \, \text{m/s} \]

3. **Use the relationship between speed of sound and temperature**:
The speed of sound in air is related to temperature by the formula:
\[ v \propto \sqrt{T} \]
Therefore, we can write:
\[ \frac{v_T}{v_{273}} = \sqrt{\frac{T}{273}} \]
where \( v_T = 336 \, \text{m/s} \) and \( v_{273} = \frac{1000}{3} \, \text{m/s} \).

4. **Substituting the known values**:
Substitute \( v_T \) and \( v_{273} \):
\[ \frac{336}{\frac{1000}{3}} = \sqrt{\frac{T}{273}} \]

5. **Simplifying the equation**:
Rearranging gives:
\[ \frac{336 \times 3}{1000} = \sqrt{\frac{T}{273}} \]
Simplifying further:
\[ \frac{1008}{1000} = \sqrt{\frac{T}{273}} \]

6. **Squaring both sides**:
\[ \left(\frac{1008}{1000}\right)^2 = \frac{T}{273} \]

7. **Calculating \( T \)**:
\[ T = 273 \times \left(\frac{1008}{1000}\right)^2 \]
\[ T = 273 \times \frac{1016064}{1000000} \]
\[ T \approx 277.41 \, \text{K} \]

8. **Convert Kelvin to Celsius**:
To convert Kelvin to Celsius:
\[ T_{Celsius} = T - 273 = 277.41 - 273 \approx 4.41^\circ C \]

### Final Answer:
The value of \( T \) is approximately \( 4.41^\circ C \).

---