Question

Two sounding sources send waves at certain temperature in air of wavelength \( 60 \text{ cm} \) and \( 60.6 \text{ cm} \) respectively. The frequency of sources differ by \( 5 \text{ Hz} \). The velocity of sound in air at same temperature is

• A. $330 \text{ m/s}$
• B. $313 \text{ m/s}$
• C. $303 \text{ m/s}$
• D. $300 \text{ m/s}$

Hint:

- Use $f = \frac{v}{\lambda}$ - Take difference carefully using reciprocals

Answer 1

The Correct Option is C

Concept: Wave relation: \[ v = f\lambda \]

Step 1: Write frequencies.
\[ f_1 = \frac{v}{\lambda_1}, \quad f_2 = \frac{v}{\lambda_2} \]

Step 2: Use given difference.
\[ |f_1 - f_2| = 5 \Rightarrow v\left|\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right| = 5 \]

Step 3: Convert wavelengths.
\[ \lambda_1 = 0.60\ \text{m}, \quad \lambda_2 = 0.606\ \text{m} \]

Step 4: Substitute values.
\[ v \left( \frac{1}{0.60} - \frac{1}{0.606} \right) = 5 \] \[ v \left( 1.6667 - 1.6502 \right) = 5 \] \[ v (0.0165) = 5 \]

Step 5: Solve for $v$.
\[ v = \frac{5}{0.0165} = 303\ \text{m/s} \]