A closed organ pipe of length L is vibrating in its first overtone. There is a point Q inside the pipe at a distance 7L/9 from the open end. The ratio of pressure amplitude at Q to the maximum pressure amplitude in the pipe is

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Question

A closed organ pipe of length L is vibrating in its first overtone. There is a point Q inside the pipe at a distance 7L/9 from the open end. The ratio of pressure amplitude at Q to the maximum pressure amplitude in the pipe is

1 : 2

2 : 1

1 : 1

2 : 3

$∆$ Pm = 2 P₀ cos kx
(assuming closed end as origin)
At point Q, $x = L - \frac{7 L}{9} = \frac{2 L}{9}$
$ΔPm = 2 {ΔP}_{0} \cos (\frac{2 π}{λ} \times \frac{2 L}{9}) = {ΔP}_{0}$
$∴$ Required ratio = 1 : 2

In a closed organ pipe (closed at one end, open at the other), the first overtone corresponds to the third harmonic. The pressure wave inside the pipe forms a standing wave. For a closed pipe, the pressure amplitude is maximum at the closed end (antinode for pressure) and zero at the open end (node for pressure). The general equation for the pressure amplitude variation along the pipe is:

$P = P_{0} \cos (k x)$ , where $k$ is the wave number and $x$ is the distance from the closed end.

Step 1: Determine the harmonic and wave number

For the first overtone (third harmonic) in a closed pipe of length L, the wavelength $λ$ is given by $λ = \frac{4 L}{3}$ . The wave number $k = \frac{2 π}{λ} = \frac{2 π}{4 L / 3} = \frac{3 π}{2 L}$ .

Step 2: Set up the coordinate system

Let the closed end be at $x = 0$ and the open end at $x = L$ . The point Q is at a distance $\frac{7 L}{9}$ from the open end, so its distance from the closed end is $x = L - \frac{7 L}{9} = \frac{2 L}{9}$ .

Step 3: Write the pressure amplitude at Q

Pressure amplitude at any point x is $P_{Q} = P_{0} \cos (k x)$ . At Q, $x = \frac{2 L}{9}$ , so:

$P_{Q} = P_{0} \cos (\frac{3 π}{2 L} \cdot \frac{2 L}{9}) = P_{0} \cos (\frac{3 π}{2} \cdot \frac{2}{9}) = P_{0} \cos (\frac{6 π}{18}) = P_{0} \cos (\frac{π}{3}) = P_{0} \cdot \frac{1}{2}$

Step 4: Find the maximum pressure amplitude

The maximum pressure amplitude occurs at the closed end (x=0) and is $P_{m a x} = P_{0}$ .

Step 5: Compute the ratio

The ratio of pressure amplitude at Q to the maximum pressure amplitude is $\frac{P_{Q}}{P_{m a x}} = \frac{P_{0} / 2}{P_{0}} = \frac{1}{2}$ . So the ratio is 1:2.

Final Answer: The ratio is 1:2.

Formulae

Wave number: $k = \frac{2 π}{λ}$

Wavelength for nth harmonic in closed pipe: $λ = \frac{4 L}{n}$ , where n is odd.

Pressure amplitude function: $P (x) = P_{0} \cos (k x)$ , with x measured from closed end.

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