Question

On a sonometer, the lengths of two wires are in the ratio \( \frac{35}{34} \), diameters are in the ratio \( \frac{4}{1} \), densities are in the ratio 1:2 and the tensions in the string are in the ratio 8:1. If the note of higher pitch has frequency of 350 Hz then find the frequency of the beats produced when sounded together:

• A. 20 Hz
• B. 7 Hz
• C. 5 Hz
• D. 10 Hz

Hint:

To calculate the beat frequency, take the difference between the two frequencies. For sonometer problems, pay attention to the ratios of the lengths, tensions, and cross-sectional areas.

Answer 1

The Correct Option is D

The frequency of a vibrating string is given by the formula:

\[ f \propto \frac{1}{L} \times \sqrt{\frac{T}{\rho A}} \] Where: - \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \rho \) is the density of the wire material,
- \( A \) is the cross-sectional area of the wire.

The frequency ratio is given by:

\[ \frac{f_2}{f_1} = \left(\frac{L_1}{L_2}\right) \times \sqrt{\frac{T_1 \rho_2 A_2}{T_2 \rho_1 A_1}} \] Substituting the given values:
- \( \frac{L_1}{L_2} = \frac{35}{34} \),
- \( \frac{T_1}{T_2} = 8 \),
- \( \frac{\rho_1}{\rho_2} = \frac{1}{2} \),
- \( \frac{A_1}{A_2} = \frac{1}{16} \),

we get:

\[ \frac{f_2}{f_1} = \frac{35}{34} \times \sqrt{\frac{8 \times 2}{16}} = \frac{35}{34} \times \sqrt{1} = \frac{35}{34} \] Therefore:

\[ f_2 = 350 \times \frac{35}{34} = 358.82 \, {Hz} \] Now, the beat frequency is:

\[ f_{{beat}} = |f_2 - f_1| = |358.82 - 350| = 8.82 \, {Hz} \approx 10 \, {Hz} \] Thus, the frequency of the beats produced is approximately 10 Hz.