Question 335: Waves

Question

The distance between two consecutive points with phase difference of $60^{\circ}$ in wave of frequency 500 Hz is 0.6 m . The velocity with which wave is travelling is

Accepted Answer

To determine the velocity of a wave given a phase difference of $60^\circ$ between two consecutive points and a frequency of 500 Hz, we follow these calculations:
Convert Phase Difference to Radians
:
$    \phi = 60^\circ = \frac{\pi}{3}    $
Relate Path Difference to Wavelength
:
The path difference $ x $ is given by:
$    x = \frac{\lambda}{2 \pi} 	imes \phi     $
where $ x = 0.6 \, 	ext{m} $.
Solve for Wavelength $\lambda$
:
$    0.6 = \frac{\lambda}{2 \pi} 	imes \frac{\pi}{3}    $
Simplifying:
$    	herefore \lambda = 3.6 \, 	ext{m}    $
Calculate Wave Velocity $v$
:
Using the relationship between velocity $v$, frequency $n$, and wavelength $\lambda$:
$    v = n \lambda = 500 	imes 3.6 = 1800 \, 	ext{m/s}    $
Convert to km/s
:
$    v = 1.8 \, 	ext{km/s}    $
Thus, the wave travels with a velocity of $1.8 \, 	ext{km/s}$.

Answer

$1.8 \mathrm{~km} / \mathrm{s}$

Answer

$9 \mathrm{~km} / \mathrm{s}$

Answer

$3.6 \mathrm{~km} / \mathrm{s}$

Answer

$2.7 \mathrm{~km} / \mathrm{s}$

Waves

The distance between two consecutive points with phase difference of 60^{\circ} in wave of frequency 500 Hz is 0.6 m . The velocity with which wave is travelling is

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