Home
Class 11
PHYSICS
(i) An organ pipe has one end closed and...

(i) An organ pipe has one end closed and at the other end there is a vibrating diaphragm. The diaphragm is a pressure node. The pipe resonates when the frequency of the diaphragm is 2 KHz. Distance between adjacent nodes is 8.0 cm. When the frequency is slowly reduced, the pipe again resonates at 1.2 KHz.
(a) Find the length of the tube.
(b) Find the next frequency above 2 KHz at which the pipe resonates.
(ii) The figure shows an arrangement for measuring the speed of sound in air. A glass tube is fitted with a movable piston that allows the indicated length L to be adjusted. There is enough gap between the piston and the tube wall to allow the air to pass through it. A speaker is placed near the open end of the tube. A microphone is placed close to the speaker and it is connected to a waveform display. The display is a pure sinusoidal waveform making 750 oscillations in 5 s. Initially, the piston is held at end A and is then slowly pulled back. Loud sound is produced by the tube when `L = 50 cm` and `L = 157 cm`. Calculate the speed of sound in air.

Promotional Banner

Similar Questions

Explore conceptually related problems

In an organ pipe the distance between the adjacent node is 4 cm . Find the frequency of source if speed of sound in air is 336 m//s .

In an organ pipe the distance between the adjacent node is 4 cm . Find the frequency of source if speed of sound in air is 336 m//s .

A tube closed at one end has a vibrating diaphragm at the other end , which may be assumed to be a displacement node . It is found that when the frequency of the diaphragm is 2000 Hz , a stationary wave pattern is set up in which the distance between adjacent nodes is 8 cm . When the frequency is gradually reduced , the stationary wave pattern reappears at a frequency of 1600 Hz . Calculate i. the speed of sound in air , ii. the distance between adjacent nodes at a frequency of 1600 Hz , iii. the distance between the diaphragm and the closed end , iv. the next lower frequencies at which stationary wave patterns will be obtained.

A tube closed at one end has a vibrating diaphragm at the other end , which may be assumed to be a displacement node . It is found that when the frequency of the diaphragm is 2000 Hz , a stationary wave pattern is set up in which the distance between adjacent nodes is 8 cm . When the frequency is gradually reduced , the stationary wave pattern reappears at a frequency of 1600 Hz . Calculate i. the speed of sound in air , ii. the distance between adjacent nodes at a frequency of 1600 Hz , iii. the distance between the diaphragm and the closed end , iv. the next lower frequencies at which stationary wave patterns will be obtained.

An organ pipe of length L is open at one end and closed at other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are

An organ pipe of length L is open at one end and closed at other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are

An organ pipe of length L is open at one end and closed at other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are

What is the shortest length of a pipe open at one end and closed to other end resonates with the fundamental frequency n ?

A string, fixed at both ends, vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6 cm. Find the length of the string.

A string, fixed at both ends, vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6 cm. Find the length of the string.