In sonometer, fundamental frequency changes from 400 Hz to 500 Hz keeping same tension. Find percentage change in length.

Question

A pipe open at both ends has a fundamental frequency \(f\) in air. The pipe is now dipped vertically in water drum to half of its length. The fundamental frequency of the air column is now equal to:

Answer 1

The fundamental frequency of a sonometer wire is given by the formula:

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

where:

f is the frequency,
L is the length of the wire,
T is the tension in the wire,
\mu is the linear mass density of the wire.

Given that the frequency changes from 400 Hz to 500 Hz while keeping the same tension, let's analyze the problem:

Let's denote the initial length of the wire as L_1 and the final length as L_2. Since the tension and mass density remain unchanged, the formula becomes:

f_1 = \frac{1}{2L_1}\sqrt{\frac{T}{\mu}} = 400 \, \text{Hz}

f_2 = \frac{1}{2L_2}\sqrt{\frac{T}{\mu}} = 500 \, \text{Hz}

Taking the ratio of the two expressions:

\frac{f_2}{f_1} = \frac{L_1}{L_2}

Substituting the given frequencies:

\frac{500}{400} = \frac{L_1}{L_2}

Simplifying the above expression:

\frac{5}{4} = \frac{L_1}{L_2}

Implying:

L_2 = \frac{4}{5}L_1

The percentage change in length is calculated using:

\text{Percentage Change} = \left(\frac{L_1 - L_2}{L_1}\right) \times 100\%

Substituting the known values:

\text{Percentage Change} = \left(\frac{L_1 - \frac{4}{5}L_1}{L_1}\right) \times 100\% = \left(\frac{1}{5}\right) \times 100\% = 20\%

Thus, the percentage change in length is 20%.

Answer 2