Question

In a string the speed of the wave is 10 $ m{s^{ - 1}} $ and its frequency is $ 100\,Hz $ . The value of the phase difference at a distance $ 2.5\,cm $ will be:
 $
  A.\,\,\dfrac{\pi }{2} \\
  B.\,\,\dfrac{\pi }{8} \\
  C.\,\,\dfrac{{3\pi }}{2} \\
  D.\,\,4\pi \\
  $

Answer 1

Hint
In the question, velocity and the frequency of the wave is given, substitute the values in the equation of the wavelength we get the value of wavelength and substituting all the known values in the equation of phase difference we get the value of phase difference at a required distance has been calculated.
The expression for finding the wavelength is
 $ \lambda = \dfrac{v}{f} $
Where, $ \lambda \, $ be the wavelength of the wave, $ v $ be the speed of the wave and $ f $ be the frequency of the wave.
The expression for finding the phase difference is
 $ \phi = \dfrac{{2\pi }}{\lambda } \times x $
Where $ x $ be the distance of the phase difference.

Complete step by step answer
From the question, Given that
Speed of the wave $ v = 10\,m{s^{ - 1}} $
Frequency of the wave $ f = 100\,Hz $
 $ \lambda = \dfrac{v}{f}.........\left( 1 \right) $
Substitute the value of $ v $ and $ f $ in the equation $ \left( 1 \right) $
 $ \lambda = \dfrac{{10}}{{100}} $
Simplify the above equation, we get
 $ \lambda = 0.1\,m $
Therefore, the wavelength of the wave, $ \lambda = 0.1\,m $
Now, we can find the value of phase difference at $ 2.5\,cm $
Phase difference $ \phi = \dfrac{{2\pi }}{\lambda } \times x.........\left( 2 \right) $
Here $ x = 2.5\,cm $
Convert the $ x $ value in terms of $ m, $ we get
 $ x = 2.5 \times {10^{ - 2}}\,m $
Substitute the value of $ x $ and $ \lambda $ in the equation $ \left( 2 \right) $
 $ \phi = \dfrac{{2\pi }}{{0.1}} \times \left( {2.5 \times {{10}^{ - 2}}} \right) $
Perform the arithmetic operations in the above equation, we get
 $ \phi = \dfrac{\pi }{2} $
Therefore, the value of the phase difference at a distance of $ 2.5\,cm $ will be $ \dfrac{\pi }{2} $ .
Hence, from the above options, the option $ \left( A \right) $ is correct.

Note
In the question, we know that the distance of the phase difference depends on the distance. If the distance value is increased or decreased the value of the phase difference is changed and given the value of the distance is in terms of $ cm $ but the value of wavelength is in terms of $ m $ . So, we convert the distance in terms of $ m. $