A) A tube is closed at one end and open at the other. It is placed in front of a

Question

A) A tube is closed at one end and open at the other. It is placed in front of a loudspeaker that is playing the sound generated by a variable frequency audio oscillator. The frequency is slowly increased from 0. The first frequency at which standing waves are generated is 325 Hz and the next is 975 Hz. The speed of sound is 340 m/s. What is the length of the tube?

B) An organ pipe, open at both ends, has successive resonances at 150 Hz and 200 Hz when the velocity of sound in air is 345 m/s. What is its length?

C)A standing wave on a string is described by y = 0.080 sin 6x cos 600t (SI units). What is the distance between successive nodes?

Explanation / Answer

A)

from the given data

first harmonic, f1 = 325Hz

third harmonic, f3 = 975 kz

so, it is a closed tube.

v = 340 m/s

let L is the length of the tube.

f1 = v/(4*L)

L = v/(4*f)

= 340/(4*325)

= 0.2615 m

B) from the given data, fundamnetal frequency, f1 = 200 - 150

= 50 hz

v = 345 m/s

apply, f1 = v/(2*L)

L = v/(2*f1)

= 345/(2*50)

= 3.45 m
C) distance between successive nodes = A/2

= 0.08/2

= 0.04 m

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