Question 412: Waves

Question

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

Accepted Answer

To determine the difference between the wavelengths of the sound waves produced by the two tuning forks, we can use the formula for the wavelength $$\lambda$$ of a wave:
$$\lambda = \frac{v}{f}$$
where $$\lambda$$ is the wavelength, $$v$$ is the velocity of sound, and $$f$$ is the frequency.
Given:
Wavelength for frequency $$320 \mathrm{~Hz}$$:
$$\lambda_1 = \frac{320 \mathrm{~ms}^{-1}}{320 \mathrm{~Hz}} = 1 \mathrm{~m}$$
Wavelength for frequency $$480 \mathrm{~Hz}$$:
$$\lambda_2 = \frac{320 \mathrm{~ms}^{-1}}{480 \mathrm{~Hz}} = \frac{2}{3} \mathrm{~m} \approx 0.67 \mathrm{~m}$$
The difference between the wavelengths is:
$$\Delta \lambda = \lambda_1 - \lambda_2 = 1 \mathrm{~m} - 0.67 \mathrm{~m} \approx 0.33 \mathrm{~m} = 33 \mathrm{~cm}$$
Therefore, the difference between the wavelengths of the waves is nearly 33 cm.
The correct option is:
Option C: 33 cm

Answer

48 cm

Answer

16.5 cm

Answer

33 cm

Answer

42 cm

Waves

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

Choose the correct answer: