A longitudinal sound wave given by `p = 2.5 sin.(pi)/(2) (x - 600 t)` (`p` is in `N//m^(2)` , x is in metal and `t` is in second) is sent down a closed a orgain pipe. If the pipe vibrates in its second overtone, the length of the pipe is

Options

A. `6 m`
B. `8 m`
C. `5 m`
D. `10 m`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the wave equation
The given wave equation is:
\[ p = 2.5 \sin\left(\frac{\pi}{2}(x - 600t)\right) \]

### Step 2: Extract the wave number (k)
In the standard form of a wave equation:
\[ p = A \sin(kx - \omega t) \]
we can identify \( k \) from the equation. Here, we have:
\[ k = \frac{\pi}{2} \]

### Step 3: Calculate the wavelength (λ)
The wavelength \( \lambda \) can be calculated using the formula:
\[ \lambda = \frac{2\pi}{k} \]
Substituting the value of \( k \):
\[ \lambda = \frac{2\pi}{\frac{\pi}{2}} = 4 \, \text{meters} \]

### Step 4: Determine the length of the pipe for the second overtone
For a closed organ pipe, the harmonics are given by:
- Fundamental frequency (1st harmonic): \( L = \frac{\lambda}{4} \)
- 1st overtone (2nd harmonic): \( L = \frac{3\lambda}{4} \)
- 2nd overtone (3rd harmonic): \( L = \frac{5\lambda}{4} \)

Since we need the length for the second overtone:
\[ L = \frac{5\lambda}{4} \]

### Step 5: Substitute the value of λ
Now substituting \( \lambda = 4 \, \text{meters} \):
\[ L = \frac{5 \times 4}{4} = 5 \, \text{meters} \]

### Conclusion
The length of the pipe is:
\[ L = 5 \, \text{meters} \]

### Final Answer
The length of the pipe is 5 meters.

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