A wave represented by the equation $y = a\cos (kx - \omega t)$ is superposed with another wave to form a stationary wave such that point $x = 0$ is a note. The equation of other wave is
\[{\text{A}}{\text{.}}\] $a\sin (kx + \omega t)$
${\text{B}}{\text{.}}$ $ - a\cos (kx + \omega t)$
${\text{C}}{\text{.}}$ $ - a\cos (kx - \omega t)$
${\text{D}}{\text{.}}$ $ - a\sin (kx - \omega t)$
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Hint: Finding the equation of the equation of a reflected wave is, write the equation of the reflected wave for a given incident wave at a rigid boundary.
They give a represented wave equation and it is from a stationary wave at some point. We find the equation of wave function.
Complete step-by-step answer:
It is stated as the question as $y = a\cos (kx - \omega t)$
Now we have to general term as, $y = a[\cos kx\cos \omega t + \sin kx\sin \omega t]$ and it is given as $y = 0{\text{ }}at{\text{ }}x = 0$ for all values of t (note), the $\cos kx$ term must vanish in the sum of the two waves.
Let us consider, ${y_1} = a\cos (kx - \omega t)$
Let the other wave equation be ${y_2} = - a\cos (kx + \omega t)$
On superposition,
$y = {y_1} + {y_2} = a\cos (kx - \omega t) - a\cos (kx + \omega t)$
Taking \[a\] as common and multiply the terms and we get,
$ \Rightarrow a[\cos kx\cos \omega t + \sin kx\sin \omega t - \cos kx\cos \omega t + \sin kx\sin \omega t]$
On cancelling the same term and adding we get
$y = 2a\sin kx\sin \omega t$
Now the given is at $x = 0$
So we get $\sin kx = 0$
$ \Rightarrow y = 0$
$\therefore x = 0$ is a note.
Hence the equation of other waves is $ - a\cos (kx - \omega t)$
So, the correct answer is option C.
Note: Our final equation describing a wave moving right is: $y(x,t) = A\sin (\omega t - kx)$ Any function where the x and $t$ dependence is of the shape $(kx - \omega t)$ represents a traveling wave of some shape.
Differential Equations:
The differential equation is one among the foremost important equations in mechanics.
Transmitted wave:
The transmitted wave is that the one moves far away from the boundary, on the opposite side of the boundary from the incident wave. Inverted means, compared to the incident wave, the disturbance within the medium is that the opposite.
They give a represented wave equation and it is from a stationary wave at some point. We find the equation of wave function.
Complete step-by-step answer:
It is stated as the question as $y = a\cos (kx - \omega t)$
Now we have to general term as, $y = a[\cos kx\cos \omega t + \sin kx\sin \omega t]$ and it is given as $y = 0{\text{ }}at{\text{ }}x = 0$ for all values of t (note), the $\cos kx$ term must vanish in the sum of the two waves.
Let us consider, ${y_1} = a\cos (kx - \omega t)$
Let the other wave equation be ${y_2} = - a\cos (kx + \omega t)$
On superposition,
$y = {y_1} + {y_2} = a\cos (kx - \omega t) - a\cos (kx + \omega t)$
Taking \[a\] as common and multiply the terms and we get,
$ \Rightarrow a[\cos kx\cos \omega t + \sin kx\sin \omega t - \cos kx\cos \omega t + \sin kx\sin \omega t]$
On cancelling the same term and adding we get
$y = 2a\sin kx\sin \omega t$
Now the given is at $x = 0$
So we get $\sin kx = 0$
$ \Rightarrow y = 0$
$\therefore x = 0$ is a note.
Hence the equation of other waves is $ - a\cos (kx - \omega t)$
So, the correct answer is option C.
Note: Our final equation describing a wave moving right is: $y(x,t) = A\sin (\omega t - kx)$ Any function where the x and $t$ dependence is of the shape $(kx - \omega t)$ represents a traveling wave of some shape.
Differential Equations:
The differential equation is one among the foremost important equations in mechanics.
Transmitted wave:
The transmitted wave is that the one moves far away from the boundary, on the opposite side of the boundary from the incident wave. Inverted means, compared to the incident wave, the disturbance within the medium is that the opposite.
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