Question

A wave of frequency 500Hz travels with a speed of 360m/s. The distance between two nearest points which are 60° out of phase is:
A. 12cm
B. 18cm
C. 50cm
D. 24cm
E. 6cm

Answer 1

Hint: The distance between two points in a wave can be found out by using the path difference. After travelling a path difference of $\lambda$ the total phase change is ${360^0}$ or $2\pi$ radians. The phase difference $\phi$ between two points and their path difference $\Delta x$ are related as $\phi = \dfrac{{2\pi }}{\lambda }\Delta x$ . Convert phase angle to radians before substituting. Do not forget to convert the obtained answer into cm.

Complete step-by-step answer:
Let’s consider the wave to be of the form $y(x,t) = Asin(wt - kx)$
Here $k = \dfrac{{2\pi }}{\lambda }$ where $\lambda$ is the wavelength of wave
And $w = 2\pi f$ where $f$ is the frequency of the wave.
This means that the particle at the origin $x = 0$ is oscillating as $Asin(wt)$
A particle at a distance $l$ from origin is oscillating as $y(l,t) = Asin(wt - kl)$
If we call $kl$ as $\phi$ , we see that the equation of motion of a particle $l$ distance away is $Asin(wt - \phi )$
The term inside the brackets () is what we call the phase of an oscillating particle. The extra term $\phi$ represents the phase difference between the two oscillations.

So now we know that a point $l$ distance away has a phase difference of $kl$ . The above derivation can also be done without taking one of the points as origin. It is true in general that the phase difference between two points separated by a distance $l$ is $kl = \dfrac{{2\pi }}{\lambda }l$ . (1)
We are asked to find the distance between two points whose phase difference is $60°$

Since $\lambda$ has to be known to apply equation (1), let’s first find $\lambda$
We know $\lambda = \dfrac{c}{f}$
Substituting the values of $c$ and $f$ from question, we get:
$\lambda = \dfrac{{360}}{{500}} = 0.72m$
Now, the phase difference between the points is given to be ${60^\circ }or\dfrac{\pi }{3}rad$
Now phase difference $\phi = \dfrac{{2\pi }}{\lambda }l$
Substituting the value of $\lambda$ and $\phi$ obtained gives :
$\dfrac{\pi }{3} = \dfrac{{2\pi }}{{0.72}}l$
$l = \dfrac{{0.72}}{6}$
$l = 0.12m = 12cm$
This is the required answer.

Note: Here we should not forget to convert from degrees to radians because the $2\pi$ in $\phi = \dfrac{{2\pi }}{\lambda }\Delta x$ is used expecting the angles to be in radians which is the total angle. If degrees are being used, $k = \dfrac{{360}}{\lambda }$ should be used. Note that this equation can also be remembered as the ratio of phase angle and total angle ( $2\pi$ ) is equal to the ratio of path difference to wavelength.
$\dfrac{\phi }{{2\pi }} = \dfrac{{\Delta x}}{\lambda }$