Question 425: Waves

Question

A sound wave is travelling with a frequency of $$50 \mathrm{~Hz}$$. The phase difference between the two points in the path of a wave is $$\frac{\pi}{3}$$. The distance between those two points is (Velocity of sound in air $$=330 \mathrm{~m} / \mathrm{s}$$ )

Accepted Answer

To find the distance between the two points in the path of a wave, we need to use the relationship between the phase difference, wavelength, and distance. The phase difference $$\Delta \phi$$ is given by:
$$\Delta \phi = \frac{2 \pi d}{\lambda}$$
where:
$$\Delta \phi$$ is the phase difference
$$d$$ is the distance between the two points
$$\lambda$$ is the wavelength
We are given:
Phase difference $$\Delta \phi = \frac{\pi}{3}$$
Velocity of sound $$v = 330 \, \mathrm{m/s}$$
Frequency $$f = 50 \, \mathrm{Hz}$$
First, we find the wavelength $$\lambda$$ using the relation:
$$\lambda = \frac{v}{f}$$
Substituting the given values:
$$\lambda = \frac{330 \, \mathrm{m/s}}{50 \, \mathrm{Hz}} = 6.6 \, \mathrm{m}$$
Now, substitute the values of $$\Delta \phi$$ and $$\lambda$$ into the phase difference formula to find $$d$$:
$$\frac{\pi}{3} = \frac{2 \pi d}{6.6}$$
Solving for $$d$$:
$$d = \frac{(\frac{\pi}{3}) \cdot 6.6}{2 \pi} = \frac{6.6}{6} = 1.1 \, \mathrm{m}$$
Thus, the distance between those two points is
1.1 m
. Therefore, the correct answer is:
Option A: 1.1 m

Answer

1.1 m

Answer

0.6 m

Answer

2.2 m

Answer

1.7 m

Waves

A sound wave is travelling with a frequency of $50 \mathrm{~Hz}. The phase difference between the two points in the path of a wave is \frac{\pi}{3}. The distance between those two points is (Velocity of sound in air =330 \mathrm{~m} / \mathrm{s}$ )

Choose the correct answer: