Maximum transverse velocity of a wave along a string

[warfreak131]

Homework Statement

The figure (attached image) shows a pulse on a string of length 100m with fixed ends. The pulse is traveling to the right without any change of shape, at a speed of 40 m/s.

What is the maximum transverse velocity of the string?

Homework Equations

y = A Sin(kx-wt)

The Attempt at a Solution

I said that the wave had a length of 2 meters. That makes the wave number 2pi/2 = pi. And since we know the velocity and wavelength, we can find the frequency, and therefore, the angular frequency. 40(m/s)/2m = 20 hz. w=2 pi f = 2 pi 20 = 40 pi. And the amplitude is .1 m.

This makes the equation y = .1 Sin(pi x - 40 pi t)

The velocity is the derivative of position, giving us -12.5 Cos[40 pi t - pi x].

Therefore, the maximum velocity would be when the derivative of velocity = 0.
y''=a= 1580 Sin[40 pi t - pi x].

I get that y'' = 0 when either x=40t or t=x/40. If I plug this into y', I get the argument in the cosine to equal 0, which makes cosine equal 1. This would just leave the constant term as the answer, making it (a magnitude of) 12.5 m/s. But according to the book, the answer is 4 m/s. How did they get this?

 

[TSny]

The graph shows a wave pulse, not a sinusoidal wave. The wave pulse would be represented by some function of the form $y(x, t) = f(x-vt)$, where $v$ is the wave speed. It will be useful to think of this as

$y(x, t) = f(u)$, where $u \equiv x - vt$

Note that at any fixed time $t$, the slope of the wave pulse at position $x$ is given by

slope = $\frac{\partial y}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x}$.

Since $\frac{\partial u}{\partial x} = 1$, we have slope = $\frac{\partial f}{\partial u}$

Similarly, the vertical velocity $V_y$ of a point of the string at position $x$ is given by

$V_y = \frac{\partial y}{\partial t} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial t}$.

Since $\frac{\partial u}{\partial t} = -v$, we have $V_y = -v \frac{\partial f}{\partial u}$.

Or, since $\frac{\partial f}{\partial u} =$ slope, $V_y = -v \cdot slope$.

You can use this to help sketch the graph of $V_y$ as a function of $x$ for the time shown in the figure.

To estimate the maximum value of $V_y \,$, use the figure to estimate the maximum slope of the wave pulse.