An open organ pipe is vibrating in its fifth overtone. The distance between two consecutive pionts where pressure amlitude is `1/sqrt 2` times pressure amplitude at pressure antinodes,is 40 cm , then the length of open organ pipe is ( neglect end connection )

Options

A. 3 cm
B. 3.6 cm
C. 4.2 cm
D. 4.8 cm

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understand the problem
We have an open organ pipe vibrating in its fifth overtone, which corresponds to the sixth harmonic. We need to find the length of the organ pipe based on the given information about pressure amplitude.

### Step 2: Determine the relationship between overtone and wavelength
For an open organ pipe, the relationship between the harmonic number (n) and the wavelength (λ) is given by:
\[ L = \frac{n \lambda}{2} \]
For the sixth harmonic (n = 6):
\[ L = \frac{6 \lambda}{2} = 3 \lambda \]

### Step 3: Analyze the pressure amplitude
The pressure amplitude varies along the length of the pipe and is given by:
\[ P(x) = P_0 \sin(kx) \]
where \( P_0 \) is the maximum pressure amplitude and \( k \) is the wave number defined as:
\[ k = \frac{2\pi}{\lambda} \]

### Step 4: Set up the equation for pressure amplitude
We are given that the pressure amplitude at certain points is \( \frac{1}{\sqrt{2}} P_0 \). Thus, we set up the equation:
\[ \sin(kx) = \frac{1}{\sqrt{2}} \]

### Step 5: Find the angles corresponding to the sine value
The sine function equals \( \frac{1}{\sqrt{2}} \) at angles:
\[ kx = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots \]
The difference between the consecutive points where this occurs is:
\[ x_2 - x_1 = \frac{3\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2} \]

### Step 6: Relate the distance to the wavelength
We know from the problem that this distance is given as 40 cm:
\[ \frac{\pi}{2} = 40 \text{ cm} \]
Converting 40 cm to meters gives:
\[ \frac{\pi}{2} = 0.4 \text{ m} \]
Thus, we can solve for \( \lambda \):
\[ \lambda = \frac{0.4 \times 2}{\pi} \]

### Step 7: Calculate the wavelength
Calculating \( \lambda \):
\[ \lambda = \frac{0.8}{\pi} \approx 0.25465 \text{ m} \]

### Step 8: Calculate the length of the organ pipe
Now substituting \( \lambda \) back into the length equation:
\[ L = 3 \lambda = 3 \times 0.25465 \approx 0.764 \text{ m} \]

### Step 9: Final answer
Thus, the length of the open organ pipe is approximately:
\[ L \approx 0.764 \text{ m} \]