Question

A pipe open only at one end has a fundamental frequency of 254 Hz. A second pipe, initially identical to the first pipe, is shortened by cutting off a portion of the open end. Now when both pipes vibrate at their fundamental frequencies, a beat frequency of 20 Hz is heard. How many centimeters were cut off the end of the second pipe? The speed of sound is 345 m/s.

Answer 1

Fundamental frequency =f1=v/4L=255

where V= velocity of sound in air

so L=V/4f1

L = 345/(4*254)

L = =0.339 m

let dL be the shortened length

then the fundamental frequency of second pipe = f2 = V/4(L-dL)

Beats=f2-f1= 20 Hz

f2 = 20 + 254 = 274 Hz

L-dL = v/4*274 = 345/(4*274) = 0.314

dL= 0.339 - 0.314

dL = 0.025 m or   2.5 cm