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
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