The Wave Equation — Question 3

Question 3

Each independent sinusoidal component must obey the same speed law

A proposed solution on a string is $y=A\cos(2x-6t)+B\cos(5x-\Omega t)$, where $A,B\ne0$ and coordinates use SI units.

1. Find the required string speed and allowed values of $\Omega$.

2. Could the second term instead have $\Omega=10\,\mathrm{rad/s}$ while the sum still satisfies one homogeneous wave equation?

Question 3 - Solution

Each independent sinusoidal component must obey the same speed law

Method and mathematical working

Each independent Fourier component must satisfy the same dispersion relation of the uniform string. A cancellation between unequal spatial harmonics cannot hold for every position.$$\begin{aligned}6^2&=v^2(2)^2\Rightarrow v=3,\\\Omega^2&=v^2(5)^2=225.\\[4pt]\Omega&=\pm15\,\mathrm{rad/s},\qquad\Omega=10:\quad25v^2-\Omega^2=225-100=125\ne0.\end{aligned}$$

(a) Calculation and conclusion

The first term requires $v^2=6^2/2^2=9$, so $v=3\,\mathrm{m/s}$. The second requires $\Omega^2=v^2(5)^2=225$, giving $\Omega=\pm15\,\mathrm{rad/s}$; the signs represent opposite directions.

(b) Interpretation and limits

No. With $\Omega=10$, the second term leaves a nonzero residual proportional to $225-100$. Since the two spatial harmonics are independent functions, a residual at one wave number cannot be canceled identically by the other. Each nonzero harmonic must satisfy the dispersion relation.

Physical check

Both signs of $\Omega$ solve the second-order wave equation; they encode opposite propagation directions for the second component. A value of $10$ does not satisfy the required coefficient relation.