The amplitude of the wave disturbance propagating in the positive x direction is given by, \[y=\dfrac{1}{(1+{{x}^{2}})}\] at time t=0 and \[y=\dfrac{1}{[1+{{(x-1)}^{2}}]}\] at t=2 seconds where x and y are in meters. The shape of wave disturbances does not change during the propagation. The velocity of the wave is:
A. 2.5 m/s
B. 0.25 m/s
C. 0.5 m/s
D. 5 m/s
Answer
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Hint: In this question we are being asked to calculate the velocity of the wave. We know that velocity is given as distance upon time. Therefore, we will first calculate the Distance travelled by the wave in a given 2 seconds instant. To do that we will find the position of wave at t=0 and t=2. It is also given that the shape of the wave does not change. Therefore, the amplitude of the wave will remain constant.
Complete answer:
It has been given that the shape of the wave does not change then as shown in the figures below the \[{{y}_{\max }}\] does not change.
It is given to us that the wave is moving in a positive X direction. Therefore, the negative amplitude of the wave will be neglected. Now let us calculate the position for the wave at t=0
It is given that at t=0
\[y=\dfrac{1}{(1+{{x}^{2}})}\]
As y is inversely proportional to x in above equation. Therefore, we know that y to be maximum the value of x will be given as
\[x=0\] ………………….. (A)
\[{{y}_{\max }}=1\] ……………………… (1)
Now, similarly for x=2
We have been given that at t=2
\[y=\dfrac{1}{[1+{{(x-1)}^{2}}]}\]
From (1) and the given condition that shape of wave does not change
We can say that
\[1=\dfrac{1}{[1+{{(x-1)}^{2}}]}\]
Therefore, on solving
We get,
\[x=1\] …………… (B)
From A and B we can say that at t=0; x=0 and at t=2; x=1 as shown in the figure above.
Therefore, we can say that the distance travelled by a wave in seconds is 1 m.
Therefore, we know velocity of wave is given by Distance upon time
Therefore, velocity is given by
\[v=\dfrac{1}{2}=0.5\]
\[v=0.5m/s\]
So, the correct answer is “Option C”.
Note:
Amplitude is defined as the maximum amount of the displacement of a particle on the medium from its position of rest. In a sense, the amplitude of the wave is the maximum distance from the rest point to the crest point in both positive and negative directions. The wavelength is measured as distance between one crest to another or one through to the other.
Complete answer:
It has been given that the shape of the wave does not change then as shown in the figures below the \[{{y}_{\max }}\] does not change.
It is given to us that the wave is moving in a positive X direction. Therefore, the negative amplitude of the wave will be neglected. Now let us calculate the position for the wave at t=0
It is given that at t=0
\[y=\dfrac{1}{(1+{{x}^{2}})}\]
As y is inversely proportional to x in above equation. Therefore, we know that y to be maximum the value of x will be given as
\[x=0\] ………………….. (A)
\[{{y}_{\max }}=1\] ……………………… (1)
Now, similarly for x=2
We have been given that at t=2
\[y=\dfrac{1}{[1+{{(x-1)}^{2}}]}\]
From (1) and the given condition that shape of wave does not change
We can say that
\[1=\dfrac{1}{[1+{{(x-1)}^{2}}]}\]
Therefore, on solving
We get,
\[x=1\] …………… (B)
From A and B we can say that at t=0; x=0 and at t=2; x=1 as shown in the figure above.
Therefore, we can say that the distance travelled by a wave in seconds is 1 m.
Therefore, we know velocity of wave is given by Distance upon time
Therefore, velocity is given by
\[v=\dfrac{1}{2}=0.5\]
\[v=0.5m/s\]
So, the correct answer is “Option C”.
Note:
Amplitude is defined as the maximum amount of the displacement of a particle on the medium from its position of rest. In a sense, the amplitude of the wave is the maximum distance from the rest point to the crest point in both positive and negative directions. The wavelength is measured as distance between one crest to another or one through to the other.
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