Problem 15.7

Young and Freedman 14th Edition.

Given:

Transverse waves on a string have a wave speed of 8.00 m/s, amplitude 0.0700 m, and wavelength 0.320 m.  The waves travel in the -x direction, and at  t = 0, the x=0 end of the string has its maximum upward displacement.

v = 8.00; A = 0.0700; λ = 0.320;

(a)  Compute related quantities f, T, k

T = λ / v

f = 1/T

ω = 2 π /T

k = 2 π / λ

(b) Wave function

The equation for the traveling wave is the following.  The ‘+‘ sign is because the wave travels to the left.  The Cos[ ] is because we are given the initial condition y[0, 0] = A.

y[x_, t_] := A Cos[k x + ω t]

Here is a snapshot of the wave at time t = 0.

Plot[y[x, 0], {x, 0, 3 λ}, 
AxesLabel -> {"t (s)", "Amp (m)"}, LabelStyle -> Larger]

Animate[Plot[y[x, t], {x, 0, 3 λ}, PlotRange->{-A, A}, PlotLabel->t,
AxesLabel -> {"t (s)", "Amp (m)"}, LabelStyle -> Larger, ImageSize -> Scaled[0.7]], {t, 0, 10 T, 0.001, AnimationRate -> 0.01, AnimationRunning -> False}, ControlPlacement -> Top]

(c).  Find the transverse displacement of a particle at x = 0.360, t = 0.150.

xc = 0.360; tc = 0.150;

y[xc, tc]

Show[{Plot[y[x,tc], {x, 0, 3λ}, PlotRange->{-A, A}, PlotLabel -> tc], 
Graphics[{Red, Disk[{xc, y[xc, tc]}, {0.02, 0.004}]}]},
AxesLabel -> {"x (m)", "Amp (m)"}, LabelStyle -> Larger, ImageSize -> Scaled[0.75]
]

(d).  Look at x = 0.360.  When does it next reach the maximum?

nf[x_] := PaddedForm[Chop[x], {4, 3}] (* Handy formatting function *)

Manipulate[Show[{Plot[y[x, t], {x, 0, 3λ}, PlotRange->{-A, A}, PlotLabel -> {nf[t], nf[y[xc,t]]}], 
Graphics[{Red, Disk[{xc, y[xc, t]}, {0.02, 0.004}]}]},
AxesLabel -> {"t (s)", "Amp (m)"}, LabelStyle -> Larger, ImageSize -> Scaled[0.75]
], {{t,0.15}, 0, 10 T, 0.001, Appearance -> "Open", AnimationRate -> 0.01}, ControlPlacement->Top]

There are many times at which y[xc] will be the maximum.  You need to tell Mathematica where to start looking.  From the animation, it appears to be about 0.155 s.

FindRoot[y[xc, t] == A, {t, tc}]

The difference from part (c) is thus about 0.005 s.

t - tc /. %

The next time y[xc] is a maximum is at t = 0.195 s.

Δt = 0.195 - 0.155

That should be one period.

T == Δt