Vibrations of a string [closed]

A string with length $L$ is stretched between two fixed points. The string can't vibrate with which wavelength?

Four options are given, namely $L$, $L/2$ , $2L$ and $4L$.

I know that the answer is $4L$ but why would a string not vibrate at that wavelength?

closed as off-topic by John Rennie, user36790, ACuriousMind♦, dmckee♦Jun 24 '16 at 16:10

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• What did you think about so far? Try to draw the four waves - or try to obtain a mathematical expression for them ... – Sanya Jun 24 '16 at 8:07
• I've already done that and still I can't understand the story behind it. – Anonymous196 Jun 24 '16 at 8:09
• Please try to add your sketches or give the mathematical expression you have arrived at, it is quite useless to just spill out the answer here without knowing where your problem really is ... – Sanya Jun 24 '16 at 8:15
• I agree with that – Anonymous196 Jun 24 '16 at 8:39
• So the mathematical expression is just to fill in lambda with $4L$. I think I know the answer because the string would become too long for it to vibrate. – Anonymous196 Jun 24 '16 at 8:42

The answer has pretty much been given in the comments, but I think a nice pictorial representation might help. The mathematical form of a standing wave is $$y(x) = \sin \left(\frac{2 \pi}{ \lambda} x \right)$$ Here $y(x)$ is the displacement of the string at point $x$. If we plot the waves for the four wavelengths we obtain the following picture I will now be pedagogical and not post the answer yet but wait for OPs comment as to why the $4 L$ one does not work.

This is of course because for the 4L wave, the fixed point at x=L is not fixed ...

To elaborate a bit on John Rennie's idea from the comments:
Where are the nodes of a wave? The nodes are where $$\frac{2 \pi}{ \lambda} x = k \pi , k \in \mathbb{Z}$$ Because this is where the sine is zero. This enables us to calculate the spacing of the nodes:

The possible values of $x$ are $\frac{ \lambda }{2} k$ for $k \in \mathbb{Z}$. The spacing of the nodes is thus $\frac{ \lambda }{2}$.

We can further think about which condition the spacing of the nodes has to fulfill for the standing wave to fit the string.

This condition is that the string length is an integer multiple of the spacing.

This way, we can also understand which of the waves cannot fit onto the string.

• That clears up a lot. What I did is make $L$ 1.0 meters long instead of $2L$. – Anonymous196 Jun 24 '16 at 9:12
• I am not quite sure I am following you - this applies for any length of string, no matter what the numerical value of L is ... – Sanya Jun 24 '16 at 9:16