Wave Speed Equation: Solve Homework w/ Max Displacement 0.16m

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SUMMARY

The wave equation y(x,t)=0.8/{(4x+5t)^2+5} describes a moving pulse with a maximum displacement of 0.16m. The pulse travels in the negative X direction with a velocity of 1.25 m/s. At t=0, the graph of the pulse is symmetric about the Y-axis, confirming its characteristics. The maximum displacement occurs at specific positions that can be calculated based on the function's parameters.

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Hydrous Caperilla
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Homework Statement



y(x,t)=0.8/{(4x+5t)^2+5 }represents a moving pulse,where x and y are in metre and tin second.Then choose the options.

(a)Pulse is moving in positive X axis
(b)In 2 secs,it will travel a displacement of 2.5m
(c)It's maximum displacement is 0.16m
(d)It is a symmetric pulse

Homework Equations

The Attempt at a Solution



So since it is a general transverse wave so it's velocity is 1.25 towards negative X axis.

I can't say about the amplitude unless I know the shape of the pulse so How should I graph this and work out the other options
 
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There is no t in the equation.
 
My mistake...sorry
 
Hydrous Caperilla said:

Homework Statement



y(x,t)=0.8/{(4x+5t)^2+5 }represents a moving pulse,where x and y are in metre and tin second.Then choose the options.

(a)Pulse is moving in positive X axis
(b)In 2 secs,it will travel a displacement of 2.5m
(c)It's maximum displacement is 0.16m
(d)It is a symmetric pulse

Homework Equations

The Attempt at a Solution



So since it is a general transverse wave so it's velocity is 1.25 towards negative X axis.

I can't say about the amplitude unless I know the shape of the pulse so How should I graph this and work out the other options
How does it look at t=0?
 
ehild said:
How does it look at t=0?
At t=0,y is 0.16m
 
Hydrous Caperilla said:
At t=0,y is 0.16m
At ##t=0##, ##y## is a function of ##x##.
 
I got the graph and it looks symmetric to Y axis with y=0.16 m at t=0
 
Hydrous Caperilla said:
I got the graph and it looks symmetric to Y axis with y=0.16 m at t=0
You should calculate the position xmax of the maximum of y (it will be a function of t), and then the value of y(xmax,t). That will tell you what the maximum displacement is. You can then check the symmetry of y around xmax.
 
Hydrous Caperilla said:
I got the graph and it looks symmetric to Y axis with y=0.16 m at t=0

Yes, it is symmetric to the Y axis at t=0. At what x does the function $$ y(x)=\frac{0.8}{(4x)^2+5} $$ have its maximum, and what is the maximum displacement from equilibrium?
At what x is the maximum of y(x,t) at a different time, at t = 2 s, for example?
 

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