A clarinet is essentially a 59.2-cm long tube that is closed at one end and open

Question

A clarinet is essentially a 59.2-cm long tube that is closed at one end and open at the other.

(a) What are the frequencies of the fundamental (lowest note the instrument can play) and the next two harmonics of a tube this length? Assume v = 343 m/s for air.

(b) Two clarinets, separated by 5.92 m, both play their lowest note at the same time. A listener sits in between them at a distance of 3.55 m from one of them. Assuming the sounds produced by the instruments are exactly in phase at the source, does the listener hear a loud sound or almost no sound at all? Explain why.

Explanation / Answer

a)frequencies of the fundamental = v/4l = 343/(4*.592) = 144.85 Hz

freq. of next two harmonics = 3v/4l = 434.54 Hz

&

5v/4l = 724.24 Hz

b)distance from other clarinet = 5.92 - 3.55 = 2.37 m

so,

time difference b/w two clarinets sound = (3.55/343) - (2.37/343) = 3.44*10^-3 sec

Time period of fundamental = 1/144.85 = 6.9*10^-3

so,time difference b/w two clarinets sound = (Time period of fundamental)/2

so, destructive interference take place

so,

listener hears almost no sound at all.

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