An open organ pipe of length l vibrates in its fundamental mode. The pressure variation is maximum

Question

An open organ pipe of length l vibrates in its fundamental mode. The pressure variation is maximum (A) At the two ends (B) At the middle of the pipe (C) At the distance (l/8) inside the ends (D) At the distance (l/4) inside the ends

Tardigrade Admin · Accepted Answer

The first normal mode of vibration is called fundamental mode.

Solution

For first normal mode of vibration

l = \frac{λ _{1}}{2}

Since, the pressure vibration is maximum at node,
Hence, the pressure variation is maximum at the

x = \frac{l}{2}

Q. An open organ pipe of length $l$ vibrates in its fundamental mode. The pressure variation is maximum

At the two ends

At the middle of the pipe

At the distance $\frac{l}{8}$ inside the ends

At the distance $\frac{l}{4}$ inside the ends

The first normal mode of vibration is called fundamental mode.

For first normal mode of vibration
$l = \frac{λ _{1}}{2}$
Since, the pressure vibration is maximum at node,
Hence, the pressure variation is maximum at the
$x = \frac{l}{2}$

Q. An open organ pipe of length l vibrates in its fundamental mode. The pressure variation is maximum

At the two ends

At the middle of the pipe

At the distance 8l​ inside the ends

At the distance 4l​ inside the ends

The first normal mode of vibration is called fundamental mode. For first normal mode of vibration l=2λ1​​ Since, the pressure vibration is maximum at node, Hence, the pressure variation is maximum at the x=2l​

Q. An open organ pipe of length $l$ vibrates in its fundamental mode. The pressure variation is maximum

At the distance $\frac{l}{8}$ inside the ends

At the distance $\frac{l}{4}$ inside the ends

The first normal mode of vibration is called fundamental mode.

For first normal mode of vibration
$l = \frac{λ _{1}}{2}$
Since, the pressure vibration is maximum at node,
Hence, the pressure variation is maximum at the
$x = \frac{l}{2}$