
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
478.2k+ views
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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

State and prove Bernoullis theorem class 11 physics CBSE

In which part of the body the blood is purified oxygenation class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE
