Question 352: Waves

Question

Velocity of sound waves in air is $330 \mathrm{~m} / \mathrm{s}$. For a particular sound wave in air, path difference of 40 cm is equivalent to phase difference of $1.6 \pi$. The frequency of this wave is

Accepted Answer

The frequency of a wave can be determined using the relationship between phase difference, path difference, and wavelength.
Given:
Phase difference, $$\Delta \phi = 1.6 \pi$$
Path difference, $$\Delta x = 40 	ext{ cm} = 0.4 	ext{ m}$$
Velocity of sound in air, $$v = 330 	ext{ m/s}$$
The phase difference is related to the path difference by the formula:
$$ \Delta \phi = \frac{2\pi}{\lambda} \Delta x $$
where $$\lambda$$ is the wavelength.
Rearranging the formula to solve for $$\lambda$$:
$$ \lambda = \frac{2\pi}{\Delta \phi} \Delta x $$
Substitute the given values:
$$  \lambda = \frac{2\pi}{1.6\pi} 	imes 0.4 	ext{ m} = \frac{2}{1.6} 	imes 0.4 	ext{ m} = 0.5 	ext{ m} $$
The frequency $$f$$ can be calculated using the wave equation:
$$ v = f \lambda $$
Solving for $$f$$:
$$ f = \frac{v}{\lambda} $$
Substitute the known values:
$$  f = \frac{330 	ext{ m/s}}{0.5 	ext{ m}} = 660 	ext{ Hz} $$
Therefore, the frequency of the wave is 660 Hz. Hence, the correct answer is:
Option C
660 Hz

Answer

165 Hz

Answer

150 Hz

Answer

660 Hz

Answer

330 Hz

Waves

Velocity of sound waves in air is 330 \mathrm{~m} / \mathrm{s}. For a particular sound wave in air, path difference of 40 cm is equivalent to phase difference of 1.6 \pi. The frequency of this wave is

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