The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of

Question

The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is 5x. The value of x is.

Accepted Answer

Given: The fifth harmonic of a closed organ pipe is in unison with the first harmonic of an open pipe.

Find: The value of $x$ if

\frac{L_c}{L_o} = \frac{5}{x}

where $L_c$ and $L_o$ are the lengths of the closed and open pipes respectively.

For a closed organ pipe, only odd harmonics are present, and the frequency of the $n$ -th harmonic is

f_n = \frac{nv}{4L_c}

So, for the fifth harmonic,

f_5 = \frac{5v}{4L_c}

For an open organ pipe, the first harmonic frequency is

f_1 = \frac{v}{2L_o}

Since the two sounds are in unison, their frequencies are equal:

\frac{5v}{4L_c} = \frac{v}{2L_o}

Cancelling $v$ and simplifying,

\frac{5}{4L_c} = \frac{1}{2L_o}

10L_o = 4L_c

\frac{L_c}{L_o} = \frac{5}{2}

Comparing with the given ratio

\frac{L_c}{L_o} = \frac{5}{x}

we get

x = 2

Therefore, the correct option is A.

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