Question

5-2 Written problem: Consider two waves on two side-by-side strings along the x-axis, one of form yi(x,t) - A sin(kx - cot) and the other of form y(x, t-A sin(kx-ω1+ φ), where A-10.0 cm, k 3.0007 cm-1 and 0.516 s, as sketched below (this is a snapshot in time - the waves are moving together to the right) Problem 5-2 What value of the phase constant ф will make y2-y/2 at t-1.5 s and at x - 0 while also making the y-velocities (the up and down string velocities) of the two waves be opposite in sign at these values of x and t? Note: these velocities are not the wave velocities. These two waves have the same k and the same o, so their wave velocities down the x-axis are exactly the same. Make sure your phase angle ф is between- and .

Answer 1

We have the 2 wave equations $y_{1}=Asin(kx-\omega t)$ and $y_{2}=Asin(kx-\omega t+\phi )$ .

At t=1.5 s and x=0 we get $y_{1}=Asin(-\omega t )=-A sin\omega t=-Asin(1.5\omega )$ and $y_2}=Asin(-\omega t+\phi )=-A sin(\omega t-\phi )=-Asin(1.5\omega-\phi )$ .

Since $y_{2}=y_{1}/2$ at this value of x and t we must have

$-Asin(1.5\omega -\phi )=-Asin(1.5\omega )/2$ i.e $sin(1.5\omega -\phi )=sin(1.5\omega )/2$

On simplifying the above equation using the trigonometric relation sin(A-B)=sin A cos B-cos A sin B we obtain:

$sin(1.5\omega )cos\phi -cos(1.5\omega )sin\phi =sin(1.5\omega )/2$ i.e

$sin(1.5\omega )(cos\phi -1/2 ) =cos(1.5\omega )sin\phi$ i.e

$tan(1.5\omega )=sin\phi /(cos\phi -1/2)=0.97$ on substituting the value of $\omega$ and determining $tan(1.5\omega )$.

i.e.$sin\phi =0.97 cos\phi -0.49$ or $sin\phi =0.97 (\sqrt{1-cos^{2}\phi })-0.49$ i.e $(sin\phi+0.49)^{2} =0.97^{2}(1-sin^{2}\phi )$ . This reduces to

$sin^{2}\phi +0.98sin\phi -0.7=0$. On solving this quadratic equation we obtain :

$sin\phi =0.48$ i.e $\phi =sin^{-1}(0.48)=0.5 rad$ or 29⁰ .

For the y velocities to be opposite in sign at these values of x and t, we must subtract an additional value of $\pi$ from the above value of $\phi$ to obtain the value of $(0.5-3.14)=-2.64 rad$ as the final value of $\phi$.