Question 337: Waves

Question

A progressive wave of frequency 400 Hz is travelling with a velocity $336 \mathrm{~m} / \mathrm{s}$. How far apart are the two points which are $60^{\circ}$ out of phase?

Accepted Answer

To find the distance between two points that are $60^{\circ}$ out of phase in a wave, it's first essential to determine the wavelength of the wave.
The wavelength ($\lambda$) can be calculated using the wave equation:
$$  v = f \lambda  $$
where:
$v$ is the velocity of the wave, $336 \, \mathrm{m/s}$,
$f$ is the frequency of the wave, $400 \, \mathrm{Hz}$.
Rearranging and solving for wavelength:
$$  \lambda = \frac{v}{f} = \frac{336 \, \mathrm{m/s}}{400 \, \mathrm{Hz}} = 0.84 \, \mathrm{m}  $$
Next, convert the phase difference from degrees to a fraction of the wavelength. The phase difference of $360^{\circ}$ corresponds to a complete cycle, i.e., one full wavelength. Therefore, $60^{\circ}$ corresponds to:
$$  	ext{Fraction of wavelength} = \frac{60^{\circ}}{360^{\circ}} = \frac{1}{6}  $$
The distance between two points that are $60^{\circ}$ out of phase is then:
$$  	ext{Distance} = \left(\frac{1}{6}ight) \lambda = \left(\frac{1}{6}ight) 	imes 0.84 \, \mathrm{m} = 0.14 \, \mathrm{m}  $$
Thus, the correct option is:
Option A
0.14 m

Answer

0.14 m

Answer

0.21 m

Answer

0.24 m

Answer

0.28 m

Waves

A progressive wave of frequency 400 Hz is travelling with a velocity 336 \mathrm{~m} / \mathrm{s}. How far apart are the two points which are 60^{\circ} out of phase?

Choose the correct answer: