Please show me step by step solution and explanation for this.
A progressing waveform along the x-axis is represented as
u(x,t) = 12 / [4 + (2x+3t+1)^4].
the unit for x and u is in m, and unit for t is seconds.
1. Find the direction of the wave and its velocity.
Hints:
The problem states "progressing waveform" and so it is the velocity of the progressing wave that must be solved. u(x,t) represents the displacement and therefore, the result of its time derivative is the velocity of what?
2. Find the amplitude of this wave.
3.From t=0 until 3 seconds later,what is the position of this wave's peak on the x-axis?
A progressing waveform along the x-axis is represented as
u(x,t) = 12 / [4 + (2x+3t+1)^4].
the unit for x and u is in m, and unit for t is seconds.
1. Find the direction of the wave and its velocity.
Hints:
The problem states "progressing waveform" and so it is the velocity of the progressing wave that must be solved. u(x,t) represents the displacement and therefore, the result of its time derivative is the velocity of what?
2. Find the amplitude of this wave.
3.From t=0 until 3 seconds later,what is the position of this wave's peak on the x-axis?
-
The displacement (u) of a wave travelling at speed v in the +x direction is:
u(x,t) = f(x – vt)
The displacement (u) of a wave travelling at speed v in the -x direction is:
u(x,t) = f(x + vt) (equation 1)
These are basic standard equations you need to know/understand. Note the coefficient in front of ‘t’ is the speed (a positive constant).
____________________
u(x,t) = 12 / [4 + (2x+3t+1)^4] (equation 2)
Let w = x + 1.5t
You can see u(x,t) = f(w) where
f(w) = 12 / [4 + (2w+1)^4]
Since u is a function of x+1.5t, by comparison with equation 1 you can see the speed is 1.5m/s and the wave travels in the –x direction (left).
So you can say velocity = -1.5m/s.
_____________________
The amplitude is the maximum value of u. From inspection you can see this is the value when the denominator of equation 2 is minimum. This will occur whenever (2x+3t+1)^4 = 0 , ie. When 2x+3t+1 = 0.
When this occurs, amplitude = u(x,t) = 12/(4+0) = 3m
_____________________
The maximum value of u (the peak’ position) occurs when 2x+3t+1 = 0.
At t=0, 2x+3*0+1=0, x= -0.5
The peak’s position is at x=-0.5m
At t=3s, 2x+3*3+1=0, x= -5
The peak’s position is at x=-5m
______________________
It is very helpful to see the shape of the wave (actually a single moving pulse).
At t=0, the shape of the pulse is given by 12 / (4+(2x+1)^4)
At t=1, the shape of the pulse is given by 12 / (4+(2x+4)^4)
You can enter each of these equations into the graph plotter at:
http://rechneronline.de/function-graphs/
and see exactly what is happening.
u(x,t) = f(x – vt)
The displacement (u) of a wave travelling at speed v in the -x direction is:
u(x,t) = f(x + vt) (equation 1)
These are basic standard equations you need to know/understand. Note the coefficient in front of ‘t’ is the speed (a positive constant).
____________________
u(x,t) = 12 / [4 + (2x+3t+1)^4] (equation 2)
Let w = x + 1.5t
You can see u(x,t) = f(w) where
f(w) = 12 / [4 + (2w+1)^4]
Since u is a function of x+1.5t, by comparison with equation 1 you can see the speed is 1.5m/s and the wave travels in the –x direction (left).
So you can say velocity = -1.5m/s.
_____________________
The amplitude is the maximum value of u. From inspection you can see this is the value when the denominator of equation 2 is minimum. This will occur whenever (2x+3t+1)^4 = 0 , ie. When 2x+3t+1 = 0.
When this occurs, amplitude = u(x,t) = 12/(4+0) = 3m
_____________________
The maximum value of u (the peak’ position) occurs when 2x+3t+1 = 0.
At t=0, 2x+3*0+1=0, x= -0.5
The peak’s position is at x=-0.5m
At t=3s, 2x+3*3+1=0, x= -5
The peak’s position is at x=-5m
______________________
It is very helpful to see the shape of the wave (actually a single moving pulse).
At t=0, the shape of the pulse is given by 12 / (4+(2x+1)^4)
At t=1, the shape of the pulse is given by 12 / (4+(2x+4)^4)
You can enter each of these equations into the graph plotter at:
http://rechneronline.de/function-graphs/
and see exactly what is happening.