The statement is as follows:

The function that describes a standing wave on a fixed string at both ends is given by $y(x, t) = 2A \sin (kx) \cos (\omega t)$, where $A = 1.5\ \mathrm{cm}$, $k = 0.3\ \mathrm{cm}^{-1}$ and $\omega = 500\ \mathrm{rad/s}$.

(a) Determine the minimum length from the rope so this standing wave can exist.
(b) Calculate the length of the string if it is vibrating at the 4th harmonic.

With the question (a) I haven't problem. I made, $k_{1}=2\pi/\lambda_{1}$ then $\lambda_{1}=20.9\ \mathrm{cm}$. Then, the minimum length that the rope can have to that the standing wave of the exercise exists is $\lambda_{1}/2=10.47\ \mathrm{cm}$

Now, regarding question (b). I know that the length of a string (I symbolize this concept with the letter $L$), is obtained with the following formula: $L=n\lambda_{n}/2$; where $n$ is the number of the harmonic. Then, $L=4\lambda_{4}/2=2v_{4}/f_{4}=2(w/k_{4})/(4f_{1})=2(w/k_{4})/[4(w/k_{1})]$ ($v_{4}$ is the velocity at the 4th harmonic). But this doesn't lead me to anything. And I can't think of another way to solve this exercise so I will be grateful for all the help possible.


closed as off-topic by G. Smith, Aaron Stevens, Gert, Buzz, John Rennie Jan 30 at 7:57

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The length given by $L$ is the length of the sting when no strain is applied to it. If you imagine a guitar string, this would be the distance between the bridge and the tuners. When you pull the string, its length is increasing. The formula you were trying to use just relates the 'relaxed' string's length and the wave length of standing wave you create.

It seems to me that in order to find the length, you essentially want to know the distance between points $a$ and $b$ along some curve $f(x)$. This is given by $$\int_{a}^{b} \sqrt{1+\left(\frac{df}{dx}\right)^2} \,dx.$$

You can read more about it here.

  • $\begingroup$ Thank you very much for the reply! One question, $A = 1, B = 4$? And who are $f$ and $x$? $\endgroup$ – Ayesca Jan 29 at 18:20
  • $\begingroup$ Sorry, $f$ is the string's wavefunction, i.e. f=y(x,t). In your example, A=0 and B=L. I edited the answer in order to reduce confusion with your notations. $\endgroup$ – Sub-Xel Jan 29 at 18:22
  • $\begingroup$ Perfect, anyway, it was very probably that was that function. Thank you! :) $\endgroup$ – Ayesca Jan 29 at 18:25

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