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.
\[\dfrac{\phi }{{2\pi }} = \dfrac{{\Delta x}}{\lambda }\]