Here is the list:

  1. While deriving the formula of fundamental frequency for a string fixed at both ends, the two waves coming from both ends are taken as $Asin(kx-wt)$ and $Asin(kx+wt+\phi).$ After applying the boundary condition that $y=0$ at $x=0$, we get that $\phi=0$. but according to my book, standing waves are obtained when waves are produced at one fixed end and then they interfere with the reflected wave. But the reflected wave is inverted, right? Doesn't this mean that the phase difference between the interferering waves if $\pi$? How can it be 0?

  2. If I take the interfering waves as $Asin(wt-kx)$ and $Asin(wt+kx+\phi)$ instead, it shouldn't make any differece, right? I've just cahnged the intitial phase. But now, after applying the boundary condition, I get $\phi=\pi$.

  3. The incoming wave is taken as Asin(kx-wt) and the reflected wave is taken as Asin(kx+wt+$\phi$). But how can we use the same 'x' in both the equations? The second equation gives the displacement at a distance x from the end and the x in first equations is measured from the origin. I'm guessing that the equatuon of the displacement produced by reflected wave should be Asin(k(L-x)+wt+$\phi$)

There's one more but it's related to sound waves:

The formula for the apparent wavelength of sound waves when the source is moving is $\lambda^{'}=\frac{v-u}{v}\lambda$ which is constant as $v$, $u$ and $\lambda$ are constants. Bu there is a diagram in my book showing the spherical wavefronts originating from a moving point source. In the diagram, the distance between successisive wavefronts decreases with time and hence wavelength varies.


If you add your two sine functions together you get a term which is position dependent and a term which is time dependent.

$$y=\sin(kx-\omega t)+ \sin(kx+\omega t + \phi) = 2\sin\left(kx+\frac \phi 2\right)\cos\left(\omega t-\frac \phi 2\right)$$

You will see from that equation you can satisfy your boundary condition by choosing an appropriate value of $\phi$.

If you choose that $y=0$ at $x=0$ then $\phi=0$ satisfies the boundary condition and $y = 2 \sin(kx)\cos(\omega t)$.
Indeed $\phi = n\pi$ where $n$ is an integer will also satisfy that boundary.

If you make $\phi = \frac \pi 2$ then $y = 2 \cos(kx)\sin(\omega t)$ and you obviously satisfy a different boundary condition.

| cite | improve this answer | |
  • $\begingroup$ I had asked two more questions. $\endgroup$ – Dove Jan 29 '17 at 10:05
  • 1
    $\begingroup$ You should make the third question a separate question as it is not related to the others. $\endgroup$ – Farcher Jan 29 '17 at 10:39
  • $\begingroup$ The $x$ in your equations is the distance of the equilibrium position of a particle from an origin. $x$ is the same in both equations. $\endgroup$ – Farcher Jan 29 '17 at 10:40
  • $\begingroup$ The second wave is coming from the end of the string. Then shouldn't its wave equation give the expression of displacent at a distance x from the end of the string? $\endgroup$ – Dove Jan 30 '17 at 3:11
  • $\begingroup$ Not at all. You have only one coordinate system and only one origin. $x$ is a position in space where waves from two sources (or one source if there is a reflection overlap). $\endgroup$ – Farcher Jan 30 '17 at 6:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.