Question

You have a stopped pipe of adjustable length close to a taut 62.0-cm, 7.25-g wire under a tension of 4710 N . You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second overtone with very large amplitude.

How long should the pipe be?

Express your answer with the appropriate units.

Answer 1

length of the wire
L = 62 cm = 0.62 m
m = 7.25 g = 0.00725 kg

linear mass density of the spting, mue = m/L

= 0.00725/0.62

= 0.0117 kg/m

T = 4710 N

speed of standing wave on the string, v = sqrt(T/mue)

= sqrt(4710/0.0117)

= 634.5 m/s

second overtone = third harmonic

= 3*fo

= 3*v/(2*L)

= 3*634.5/(2*0.62)

= 1535 Hz

we know, velocityb of sound, vo = 343 m/s

let l is the length of the stopped pipe.

fundamental frequency of stopped pipe = v/(4*l)

1535 = 343/(4*l)

l = 343/(4*1535)

= 0.0559 m or 5.59 cm <<<<<<<<--------------Answer