Exercise about the wavefunction

AI Thread Summary
The discussion revolves around finding the wavefunction for a wave on an ideal rope and determining the distance between two points displaced from equilibrium. The wavefunction is derived as ξ(x,t) = 0.1 sin(6t - 0.3x), with parameters identified as ξ0 = -0.1 m, ω = 6 rad/s, and ψ = 0. The issue arises in calculating the distance δ between points where the displacement is ±0.02 m, with the initial approach focusing only on positive displacement. To resolve the problem, it is suggested to consider both positive and negative displacements and visualize the sine wave to accurately identify the points on the x-axis. Understanding these aspects will lead to the correct calculation of the distance δ.
Soren4
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Homework Statement


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Consider an ideal rope where there is a wave moving at velocity ##v=20 m/s##. The displacement of one end of the rope is given by
$$s(t)=0.1 \mathrm{sin}(6 t)$$
a) Find the wavefunction ##\xi(x,t)##, knowing that it is progressive
b) Find the distance ##\delta## (in absolute value) between two points of the rope, that, at a certain time istant, are displaced from Equilibrium position of ##0.02 \mathrm{m}##
##[\mathrm{Result} \, \delta=1.34 m]##

Homework Equations


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A general wavefunction is
$$\xi(x,t)=\xi_0 \mathrm{sin}( k x-\omega t +\psi)\tag{1}$$

The Attempt at a Solution


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a) My boundary condition for ##(1)## is that
$$\xi(0,t)=0.1 \mathrm{sin}(6 t)\tag{2}$$
Now is it correct to conclude that ##\xi_0=-0.1 m##, ##\omega=6 rad/s## and ##\psi=0##?
If so, then, considering also ##k=\frac{2 \pi}{\lambda}=\frac{\omega}{v}=0.3 \frac{1}{m}##
$$\xi(x,t)=0.1 \mathrm{sin}(6t-0.3x )\tag{3}$$

b)Here is the problem. I would say that

$$0.02=0.1 \mathrm{sin}(6t-0.3x )\implies (6t-0.3x)=\arcsin(0.2)+2n\pi \vee \pi-\arcsin(0.2)+2n\pi \implies |x_2-x_1|=\frac{\pi-2\arcsin(0.2)}{0.3}$$

But this does not give the correct result.

Where did I go wrong in this problem? Any suggestion is highly apprectiated.
 
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Apparently they want the smallest Δx between two points for which the absolute value of the displacement ##\xi## is 0.02 m.
 
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Likes Soren4
Thanks for the reply!

Yes that's part of the question I guess, could you suggest where I went wrong in my derivation? I still do not see the mistake..
 
It looks like your solution only considers values of ##x## where ##\xi = +0.02## m. What about points where ##\xi = -0.02## m? It helps to make a sketch of the sine wave at some instant of time and mark the points on the x-axis where ##\xi = \pm 0.02## m.
 
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