Wave equation and graphing wave instance

[sapiental]

If the end of a string is given a single shake, a wave pulse
propagates down the string. A particular wave pulse is described by the function

y(x,t) = (A^3/(A^2 + (x - vt)^2))


where A = 1.00 cm, and v = 20.0 m/s.
a) Sketch the pulse as a function of x at t = 0. How far along the string does the pulse
extend?
b) Sketch the pulse as a function of x at t = 0.001 s.
c) At the point x = 4.50 cm, at what time t is the displacement maximum?
d) At which two times is the displacement at x = 4.50 cm equal to half its maximum value?
e) Show that y(x, t) satisfies the wave equation.


a) y(x, t = 0) = (A^3/(A^2 + (x - v(0))^2))

i used the command plot(x,y) I let x = 0 through 1 (this is MATLAB a simple graphing program)

plot(x, (.01m^3/(.01m^2 + (x)^2)

(Please see my attached graph a)

b) y(x, t = .001) = (A^3/(A^2 + (x - vt)^2))

Please see my attached graph1

c) y(x = .045m, t) = (A^3/(A^2 + (x - vt)^2))

I want to graph this as a function of t( t being the x axis) what range should I use for t ( i.e 1s, 2s..)?

My attempt is to let t = 0:.1:1 making the x-axis values (0s, .1s, .2s, ... 1s)

so to plot t vs (A^3/(A^2 + (x - vt)^2))

d) Can someone give me an iedea of how to graph the equation to get the actual wave on mt graph?

e) I think I have to take the partial derivative of either the wave equation or this particular pulse equation but I'm very bad at calc and havn't covered partial derivatives. If somebody could do this out for me step by step I would really learn alot.

 

[Jhero]

a) The pulse extends along the string until it reaches the end, so it extends infinitely in the x direction. However, its amplitude decreases as it propagates down the string.

b) At t = 0.001 s, the pulse will have moved a distance of 20 cm (20 m/s * 0.001 s) along the string. So the pulse will be centered at x = 20 cm, and its amplitude will be smaller compared to the pulse at t = 0.

c) To find the time at which the displacement is maximum at x = 4.50 cm, we need to set x = 4.50 cm and solve for t in the equation y(x, t) = (A^3/(A^2 + (x - vt)^2)) = A^3/(A^2 + (4.50 cm - 20 m/s * t)^2). This gives t = 0.225 s.

d) We can find the times at which the displacement at x = 4.50 cm is half its maximum value by setting y(x, t) = 0.5 * A^3/(A^2 + (4.50 cm - 20 m/s * t)^2) and solving for t. This gives two solutions: t = 0.1125 s and t = 0.3375 s.

e) To show that y(x, t) satisfies the wave equation, we need to take the partial derivatives of y with respect to x and t and then substitute them into the wave equation, which is ∂^2y/∂x^2 = (1/v^2) * ∂^2y/∂t^2. This will show that the second derivatives are equal, thus proving that y(x, t) satisfies the wave equation.