Question

Two waves, each having a frequency of 100 Hz and a wavelength of 2⋅0 cm, are travelling in the same direction on a string. What is the phase difference between the waves (a) if the second wave was produced 0⋅015 s later than the first one at the same place, (b) if the two waves were produced at the same instant but first one was produced a distance 4⋅0 cm behind the second one? (c) If each of the waves has an amplitude of 2⋅0 mm, what would be the amplitudes of the resultant waves in part (a) and (b) ?

Answer 1

Given:
Two waves have same frequency (f), which is 100 Hz.
Wavelength (λ) = 2.0 cm
\[= 2 \times  {10}^{- 2} m\]

Wave  speed,  \[\nu   = f \times \lambda = 100 \times 2 \times  {10}^{- 2}   m/s\] 

\[  = 2  m/s\]
(a) First wave will travel the distance in 0.015 s.

\[\Rightarrow x = 0 . 015 \times 2\] 

\[     = 0 . 03  m\]
This will be the path difference between the two waves.
So, the corresponding phase difference will be as follows:

\[\phi = \frac{2\pi  x}{\lambda}\] 

\[= \left\{ \frac{2\pi}{\left( 2 \times {10}^{- 2} \right)} \right\} \times 0 . 03 = 3\pi\]

(b) Path difference between the two waves, x = 4 cm = 0.04 m
So, the corresponding phase difference will be as follows:

\[\Rightarrow \phi = \frac{2\pi x}{\lambda}  \] 

\[ = \left\{ \frac{2\pi}{\left( 2 \times {10}^{- 2} \right)} \times 0 . 04 \right\}\] 

\[= 4\pi\]
(c) The waves have same frequency, same wavelength and same amplitude.
Let the wave equation for the two waves be as follows:

\[y_1  = r  \sin  \omega t\] 

\[And,    y_2  = r  \sin  \left( \omega t + \phi \right)\] 

\[\text{ By  using  the  principle  of  superposition: }\] 

\[y =  y_1  +  y_2 \] 

\[     = r\left[ \sin\omega t + \left( \omega t + \phi \right) \right]\] 

\[       = 2r\sin\left( \omega t + \frac{\phi}{2} \right)  \cos\left( \frac{\phi}{2} \right)\]
∴ Resultant amplitude
\[= 2r  \cos\frac{\phi}{2}\]

\[So,   \text{ when }  \phi = 3x: \] 

\[ \Rightarrow r = 2 \times  {10}^{- 3}   m\] 

\[ R_{resultant}  = 2 \times \left( 2 \times {10}^{- 3} \right)  \cos  \left( \frac{3\pi}{2} \right)\] 

\[ = 0\]

\[Again,   when  \phi = 4\pi: \] 

\[ R_{resultant}  = 2 \times \left( 2 \times {10}^{- 3} \right)  \cos  \left( \frac{4\pi}{2} \right)\] 

\[ = 4 \times  {10}^{- 3}  \times 1  \] 

\[ = 4  mm\]