
A travelling wave pulse is given by \[y=\dfrac{10}{5+{{(x+2t)}^{2}}}\]. Here, x and y are in metre and t in second. In which direction and with what velocity is the pulse propagating? What is the amplitude of pulse?
Answer
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Hint: We are given with the equation of a travelling wave pulse, although this form is not the standard but we can compare this with the standard equation of a travelling wave pulse to find out the velocity and the direction of propagation of the wave pulse. Also, in order to find out the amplitude, we can use the differential.
Complete step by step answer:
Given, \[y=\dfrac{10}{5+{{(x+2t)}^{2}}}\], comparing it with the standard equation of the wave, \[y=f(x-vt)\] we get
V= 2 m/s and it is moving in negative x direction.
Also, amplitude is the maximum displacement, \[\therefore \dfrac{dy}{dt}=0\]
$\dfrac{d[\dfrac{10}{5+{{(x+2t)}^{2}}}]}{dt}=0 \\
\Rightarrow -10{{[5+{{(x+2t)}^{2}}]}^{2}}(2)(2)(x+2t)=0 \\
\Rightarrow x+2t=0 \\
\therefore t=\dfrac{-x}{2} \\$
So, the equation becomes,
${{y}_{\max }}=\dfrac{10}{5+{{(x+2\times \dfrac{-x}{2})}^{2}}} \\
\Rightarrow {{y}_{\max }}=\dfrac{10}{5} \\
\therefore {{y}_{\max }}=2 $
The amplitude of the pulse is 2m.
Additional Information:
Standing waves are produced when waves travel in diametrically opposite directions. Nodes and antinodes are a region where there are no vibrations. The points in a standing wave that appear to remain flat and do not move are called nodes. Wave is reflected and reflection a node is formed. We know when a stationary wave is produced the fixed ends behave as a node.
Note:While solving such kinds of problems, the easiest way to tackle them is to compare the given equation to the standard equation. Sometimes, the equation given in the question is a bit complex, then we have to simplify it and modify it into a form so that we can compare it with the standard equation. Amplitude is the maximum displacement.
Complete step by step answer:
Given, \[y=\dfrac{10}{5+{{(x+2t)}^{2}}}\], comparing it with the standard equation of the wave, \[y=f(x-vt)\] we get
V= 2 m/s and it is moving in negative x direction.
Also, amplitude is the maximum displacement, \[\therefore \dfrac{dy}{dt}=0\]
$\dfrac{d[\dfrac{10}{5+{{(x+2t)}^{2}}}]}{dt}=0 \\
\Rightarrow -10{{[5+{{(x+2t)}^{2}}]}^{2}}(2)(2)(x+2t)=0 \\
\Rightarrow x+2t=0 \\
\therefore t=\dfrac{-x}{2} \\$
So, the equation becomes,
${{y}_{\max }}=\dfrac{10}{5+{{(x+2\times \dfrac{-x}{2})}^{2}}} \\
\Rightarrow {{y}_{\max }}=\dfrac{10}{5} \\
\therefore {{y}_{\max }}=2 $
The amplitude of the pulse is 2m.
Additional Information:
Standing waves are produced when waves travel in diametrically opposite directions. Nodes and antinodes are a region where there are no vibrations. The points in a standing wave that appear to remain flat and do not move are called nodes. Wave is reflected and reflection a node is formed. We know when a stationary wave is produced the fixed ends behave as a node.
Note:While solving such kinds of problems, the easiest way to tackle them is to compare the given equation to the standard equation. Sometimes, the equation given in the question is a bit complex, then we have to simplify it and modify it into a form so that we can compare it with the standard equation. Amplitude is the maximum displacement.
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