Question 282: Waves

Question

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

Accepted Answer

Given:
Frequency, $f = 400 ~	ext{Hz}$
Velocity, $v = 336 ~	ext{m/s}$
Phase difference, $\Delta\phi = 60^\circ$
Step 1: Calculate the wavelength $(\lambda)$.
We know,
$$ v = f \lambda $$
So,
$$ \lambda = \frac{v}{f} = \frac{336}{400} = 0.84~	ext{m} $$
Step 2: Relate the phase difference to distance $(x)$.
A phase difference $360^\circ$ (or $2\pi$ radians) corresponds to one full wavelength:
$$ \frac{\Delta\phi}{360^\circ} = \frac{x}{\lambda} $$
Substitute the values:
$\Delta\phi = 60^\circ$
$\lambda = 0.84~	ext{m}$
So,
$$ \frac{60^\circ}{360^\circ} = \frac{x}{0.84} $$
$$ \frac{1}{6} = \frac{x}{0.84} $$
$$ x = \frac{0.84}{6} = 0.14~	ext{m} $$
Final Answer:
$ \boxed{0.14~	ext{m}} $
Correct option: Option B

Answer

0.12 m

Answer

0.14 m

Answer

0.21 m

Answer

0.28 m

Waves

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

Choose the correct answer: