Question

A wave of frequency \[{\rm{500Hz}}\] has a velocity of \[{\rm{360m}}{{\rm{s}}^{{\rm{ - 1}}}}\]. Calculate the distance between two points that are \[{60^0}\] out of phase.
A. 12 cm
B. 18 cm
C. 17 cm
D. 25 cm

Answer 1

Hint:To proceed with the problem, let’s see what the path difference and phase difference say. The distance travelled by the two waves from their respective sources to a given point on the pattern is known as path difference. The phase difference is defined as the difference in phase angle between two waves. Now we can solve the problem by considering the definitions.

Formula used:
The formula to find the wavelength is,
\[\lambda = \dfrac{v}{f}\]
Where, \[v\] is the velocity of the wave and \[f\] is the frequency of the wave.
To find the path difference, the formula is given by,
\[\Delta x = \dfrac{\lambda }{{2\pi }}\Delta \varphi \]
Where, \[\Delta x\] is the path difference, \[\lambda \] is the wavelength and \[\Delta \varphi \] is the phase difference.

Complete step by step solution:
To find the path difference, first, we need to find the
\[\lambda = \dfrac{v}{f}\]
\[ \Rightarrow \lambda = \dfrac{{360}}{{500}}\]
By data we have \[{\rm{v = 360m}}{{\rm{s}}^{{\rm{ - 1}}}}\] and \[f = 500Hz\].
\[\lambda = 0.72\,m\]

Now let’s find the path difference or the distance between the two points from the above formula,
\[\Delta x = \dfrac{\lambda }{{2\pi }}\Delta \varphi \]
\[ \Rightarrow \Delta x = \dfrac{\lambda }{{2\pi }} \times \dfrac{\pi }{3}\]
By data, \[\Delta \varphi = {60^0} = \dfrac{\pi }{3}\]
\[ \Rightarrow \Delta x = \dfrac{\lambda }{6}\]

Now substitute the value of \[\lambda \]in the above equation we get,
\[\Delta x = \dfrac{{0.72}}{6}\]
\[ \Rightarrow \Delta x = 0.12\,m\]
Convert the above value from m to cm we get,
\[\therefore \Delta x = 12\,cm\]
Therefore, the distance between the two points is 12 cm.

Hence, option A is the correct answer.

Note:Now will see on what factors the frequency of a wave depends. As we know that the frequency is the number of cycles per second hence it depends only on the frequency of the source.