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) Write out a function for the wave y(x, t)

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], {t, 0, 10 T, 0.001, AnimationRate -> 0.01, AnimationRunning -> False}]

(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 that same x-position, namely (Subscript[x, c]) = 0.360.  When does it next reach the maximum?

There are two approaches to take.  The first is essentially graphical -- draw graphs of the function and look at the clock, waiting for the height at (Subscript[x, c]) to reach a maximum.  The second is a more analytical approach.

Graphical and numerical approach

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[xc], 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}]

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.  (Tell Mathematica to start looking at 0.18 s, that is, at some time significantly later than 0.155 s.)

FindRoot[y[xc, t] == A, {t, 0.18}]

Compute how much later that is than the previous maximum.  It should be one period.

Δt = 0.195 - 0.155

T == Δt

Analytic approach

The problem is to solve for the time when y((Subscript[x, c]), t) = A, that is, the maximum.  The position is the same as in part (c), but the time is unknown.
	y((Subscript[x, c]) , t) = A cos(k (Subscript[x, c]) + ω t)
A = A cos(k (Subscript[x, c]) + ω t)
1 =  cos(k (Subscript[x, c]) + ω t)
The cosine function is 1 whenever the argument is an integer number of 2π (e.g. 2π, 4 π, 6π, etc.).  Write that as n*2π, where “n” is an integer.
	cos(k (Subscript[x, c]) + ω t)  = 1
(k (Subscript[x, c]) + ω t) = n * 2π
t = (1/ω)(n * 2π - k (Subscript[x, c]))
We don’t know which ‘n’ to pick however.  We do know that t must be greater than the time in part (c), which was 0.15 s (because the problem asked for “the next time”).  So we can just keep trying ‘n’ values until the answer is greater than 0.15 s.

{ω, k, xc} (* Reminder of these specific numerical values *)

(1/ω)(1 * 2π - k xc) (* n = 1 *)

(1/ω)(2* 2π - k xc) (* n = 2 *)

(1/ω)(3 * 2π - k xc) (* n = 3 *)

(1/ω)(4 * 2π - k xc) (* n = 4 *)

(1/ω)(5 * 2π - k xc) (* n = 5 *)

The time 0.155 seconds is the first value after part (c).  It corresponds to the integer ‘n’ = 5.