Question

When a metal pipe is cut into two pieces, the lowest resonant frequency for one piece is 256 Hz and for the other is 440 Hz. What resonant frequency would have been produced by the original pipe? What was the length of the original pipe?

Answer 1

Let the orizinal length be L

let it is cut into a and L-a length

Original pipe f = v/2L

After cut

f1 = v/2*a

2a/v = 1/f1 ---------------(1)

f2 = v/2(L-a)

1/f2 = 2(L-a)/v = 2L/v - 2a/v

From eq(1) , put the value , we get

1/f2 = 1/f - 1/f1

f = f1f2/(f1 + f2)

= (256*440)/(256+440)

= 161.839 Hz

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Therefore

v/2L = 161.839

343/(2L) = 161.839

L = 1.0596 m

Answer 2

L1=$\lambda$/2

L1=$\lambda$/2

frequency=Vsound/2*L1=256Hz

L1=Vsound/512

L2=$\lambda$/2

L2=$\lambda$/2

frequency=Vsound/2*L2=440Hz

L2=vsound/880

total length L=L1+L2

resonant frequency of total pipe=Vsound/2L

=Vsound/2*(L1+L2)

=Vsound/2*(Vsound/512+Vsound/880)

=162.003Hz

total length=Velocity of sound/324.006m