Question

Two tuning forks of frequencies $320\text{ Hz}$ and $480\text{ Hz}$ are sounded together to produce sound waves. The velocity of sound in air is $320\text{ ms}^{-1}$. The difference between wavelengths of these waves is nearly

• A. $48\text{ cm}$
• B. $16.5\text{ cm}$
• C. $33\text{ cm}$
• D. $42\text{ cm}$

Hint:

When the velocity match exactly equals one of the frequencies ($v = f_1 = 320$), its matching wavelength calculates instantly to a perfect unit integer ($1\text{ m}$). This trivial cancellation allows you to immediately pinpoint the subtraction as $1 - \frac{2}{3} = \frac{1}{3}\text{ m} \approx 33\text{ cm}$.

Answer 1

The Correct Option is C

Step 1: Recognise common speed, differing frequencies.
Both forks share the same sound speed in air, so wavelength is just $\lambda = v/f$. We compute each wavelength and subtract.
Step 2: Wavelength of the lower fork.
$\lambda_1 = \dfrac{v}{f_1} = \dfrac{320}{320} = 1\,\text{m}$.
Step 3: Wavelength of the higher fork.
$\lambda_2 = \dfrac{v}{f_2} = \dfrac{320}{480} = \dfrac{2}{3}\,\text{m} \approx 0.667\,\text{m}$.
Step 4: Take the difference.
$\Delta\lambda = 1 - \dfrac{2}{3} = \dfrac{1}{3}\,\text{m}$.
Step 5: Convert to centimetres.
$\dfrac{1}{3}\,\text{m} = 33.3\,\text{cm}$.
Step 6: Conclude.
The wavelength difference is nearly $33\,\text{cm}$. \[ \boxed{\Delta\lambda \approx 33\ \text{cm}} \]