A wave travelling along a positive $ x - axis $ is given by $ y\; = Asin\left( {\omega t - kx} \right) $ . If it is reflected from a rigid boundary such that 80% amplitude is reflected, then equation of reflected wave is
(A) $ y\; = Asin\left( {\omega t - 0.8kx} \right) $
(B) $ y\; = - 0.8Asin\left( {\omega t + kx} \right) $
(C) $ y\; = Asin\left( {\omega t + kx} \right) $
(D) $ y\; = 0.8Asin\left( {\omega t - kx} \right) $
Answer
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Hint
The reflected wave has an amplitude of 0.8 times that of the original wave because only 80% of the amplitude is being reflected. After getting reflected the wave travels in the negative x-axis direction. The phase change is given as $ \pi = 180^\circ $ , which changes the sign of the wave equation because $ \sin (\theta + \pi ) = - sin(\theta ) $.
Complete step by step answer
The equation of the original wave is given as $ y\; = Asin\left( {\omega t - kx} \right) $
Only $ 80\% $ of the amplitude is reflected so the equation of reflected wave changes to
$ y\; = 0.8Asin\left( {\omega t - kx} \right) $
After reflection the wave travels in the negative $ x - axis $ direction, we represent this in the equation by
$ y\; = 0.8Asin\left( {\omega t - k( - x)} \right)\; = 0.8Asin\left( {\omega t + kx} \right) $
The final condition,
Phase change is given as $ \pi $ from trigonometry $ \sin (\theta + \pi ) = - sin(\theta ) $
$ y\; = - 0.8Asin\left( {\omega t + kx} \right) $
The final equation if the reflected wave is given as $ y\; = - 0.8Asin\left( {\omega t + kx} \right) $
Hence option (B) $ y\; = - 0.8Asin\left( {\omega t + kx} \right) $ is the correct answer.
Additional Information
Reflection involves a change in direction of waves when they bounce off a barrier. The refraction of waves involves a change in the direction of waves as they pass from one medium to another. Refraction, or the bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves.
Note
Students might leave the question without changing the sign of the equation due to phase change. Always remember to do so. The wave equation is an important second-order linear partial differential equation for the description of waves
The reflected wave has an amplitude of 0.8 times that of the original wave because only 80% of the amplitude is being reflected. After getting reflected the wave travels in the negative x-axis direction. The phase change is given as $ \pi = 180^\circ $ , which changes the sign of the wave equation because $ \sin (\theta + \pi ) = - sin(\theta ) $.
Complete step by step answer
The equation of the original wave is given as $ y\; = Asin\left( {\omega t - kx} \right) $
Only $ 80\% $ of the amplitude is reflected so the equation of reflected wave changes to
$ y\; = 0.8Asin\left( {\omega t - kx} \right) $
After reflection the wave travels in the negative $ x - axis $ direction, we represent this in the equation by
$ y\; = 0.8Asin\left( {\omega t - k( - x)} \right)\; = 0.8Asin\left( {\omega t + kx} \right) $
The final condition,
Phase change is given as $ \pi $ from trigonometry $ \sin (\theta + \pi ) = - sin(\theta ) $
$ y\; = - 0.8Asin\left( {\omega t + kx} \right) $
The final equation if the reflected wave is given as $ y\; = - 0.8Asin\left( {\omega t + kx} \right) $
Hence option (B) $ y\; = - 0.8Asin\left( {\omega t + kx} \right) $ is the correct answer.
Additional Information
Reflection involves a change in direction of waves when they bounce off a barrier. The refraction of waves involves a change in the direction of waves as they pass from one medium to another. Refraction, or the bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves.
Note
Students might leave the question without changing the sign of the equation due to phase change. Always remember to do so. The wave equation is an important second-order linear partial differential equation for the description of waves
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