The Wave Equation — Question 8

Section 16-7 · Physics pages 2526–2527

Question 8 A forcing term can be recovered from a failed homogeneous solution A uniform string obeys µytt = Fyxx + p(x, t), where p is an externally applied transverse force per unit length. A prescribed motion is y = A cos(kx) cos Ωt. x y p(x, t) A distributed transverse force maintains the prescribed cosine shape; no endpoint constraints are implied. (a) Find the required force density. (b) For Ω= vk with v2 = F/µ, what forcing is required in the interior? Why might an experiment still require a driver?
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Question 8 - Solution A forcing term can be recovered from a failed homogeneous solution x y p(x, t) Method and mathematical working Move the inertial and tension terms to the side defining the applied force per length. Evaluate both second derivatives of the prescribed standing profile. p = µytt −Fyxx, ytt = −Ω2y, yxx = −k2y, p = (Fk2 −µΩ2)y. p = µ(v2k2 −Ω2)y, Ω= vk ⇒p = 0. (a) Calculation and conclusion ytt = −Ω2y and yxx = −k2y, so p = µytt −Fyxx = (Fk2 −µΩ2)y. Its units are force per unit length. (b) Interpretation and limits At Ω= vk, the interior forcing vanishes: this is a free solution if boundaries and initial conditions are compatible. A real system can still need a driver to replace dissipative losses or establish the motion. The ideal homogeneous equation contains no such losses. Physical check When Ω2 = Fk2/µ, the required interior forcing vanishes. Endpoint compatibility and the initial state still determine whether that unforced motion is physically realized.

Question 8

A forcing term can be recovered from a failed homogeneous solution

A uniform string obeys μytt=Fyxx+p(x,t)\mu y_{tt}=F y_{xx}+p(x,t), where pp is an externally applied transverse force per unit length. A prescribed motion is y=Acos⁡(kx)cos⁡Ωty=A\cos(kx)\cos\Omega t.

Diagram shown in the original SVG worksheet.

  1. Find the required force density.

  2. For Ω=vk\Omega=vk with v2=F/μv^2=F/\mu, what forcing is required in the interior? Why might an experiment still require a driver?

Question 8 - Solution

A forcing term can be recovered from a failed homogeneous solution

Diagram shown in the original SVG worksheet.

Method and mathematical working

Move the inertial and tension terms to the side defining the applied force per length. Evaluate both second derivatives of the prescribed standing profile.p=μytt−Fyxx,ytt=−Ω2y,yxx=−k2y,p=(Fk2−μΩ2)y.p=μ(v2k2−Ω2)y,Ω=vk⇒p=0.\begin{aligned}p&=\mu y_{tt}-F y_{xx},\\y_{tt}&=-\Omega^2y,\qquad y_{xx}=-k^2y,\\p&=(Fk^2-\mu\Omega^2)y.\\[4pt]p&=\mu(v^2k^2-\Omega^2)y,\qquad\Omega=vk\Rightarrow p=0.\end{aligned}

(a) Calculation and conclusion

ytt=−Ω2yy_{tt}=-\Omega^2y and yxx=−k2yy_{xx}=-k^2y, so p=μytt−Fyxx=(Fk2−μΩ2)yp=\mu y_{tt}-F y_{xx}=(Fk^2-\mu\Omega^2)y. Its units are force per unit length.

(b) Interpretation and limits

At Ω=vk\Omega=vk, the interior forcing vanishes: this is a free solution if boundaries and initial conditions are compatible. A real system can still need a driver to replace dissipative losses or establish the motion. The ideal homogeneous equation contains no such losses.

Physical check

When Ω2=Fk2/μ\Omega^2=Fk^2/\mu, the required interior forcing vanishes. Endpoint compatibility and the initial state still determine whether that unforced motion is physically realized.