I'm reading all over the internet that there is no relationship between frequency and amplitude of a wave given the context of a simple experiment where a string is tied between two points and one of the points is made to oscillate with some frequency $f$. However, if we solve the first equation for amplitude we see that

$$ \frac {y(x,t)}{\sin(kx-wt)} = A $$

And knowing that $w = 2\pi f$, couldn't we say that there is at least some sort of relationship between amplitude and frequency?

  • $\begingroup$ Your text is almost certainly using a lower-case omega ($\omega$) not a latin double-u ($w$) for the angular frequency. $\endgroup$ Jul 4, 2015 at 15:28

1 Answer 1


In the equation:

$$y(x,t)=Asin ~k(x-vt)$$

$A$ can be varied independently of $k$ and $v$ and hence of $f$. That is what is meant by saying that the amplitude doesn't depend on the frequency. Now, when you write the equation as:

$$A = \frac{y(x,t)}{sin ~k(x-vt)}$$

it means that the ratio of the height of the string from the mean position at some point to a function($sin ~k(x-vt)$) is always constant and equal to the amplitude. So it is basically saying that the height of the string at such $(x,t)$ where $k(x-vt) = (2n+\frac{1}{2})\pi$ is equal to the amplitude. But does this mean that you can change the ratio by changing $k$ and $v$? No you can't. And why is that? Because the numerator $y(x,t)$ also depends on $k$ and $v$.

In other words, when writing your equation for $A$ you are wrongly assuming that you can change the $k$ and $v$ in the denominator without changing the numerator. Whatever changes you make in the denominator will be reflected in the numerator and hence the ratio will remain fixed. That indeed is what the equation is saying- no matter what your $k$ and $v$ are, take this ratio and you get the amplitude.

Attempting to give more physical intuition. Imagine that you have a vibrating string and you need to find its amplitude. At what point and time will you measure its height to ensure that you got the amplitude? The crest of the wave will keep moving, making it hard to take readings.

So imagine you get another string also tied at both ends and exactly identical to the first, except that its amplitude is always 1 (which I am not saying is trivial to set up, but well this is a thought experiment). So now you just go ahead and measure the heights of both the strings at the same point simultaneously and take a ratio of the heights. You get the amplitude of the string, just as your equation says.

Now, how will you go about changing the $k$ or $v$ or $\omega$ in one string (the unit amplitude one) without changing them in the other? You have to use the same driving source so that the strings are phase matched so you cant vary the frequency of one without changing the other as well. Since they are identical you cannot change the material of one without the other. So you cannot change the denominator of your equation without also changing the numerator simultaneously and canceling out the change.

And finally a short technical answer that will make perfect sense if you are mathematically oriented.

Notice that $y(x,t)=Asin(kx-wt)$ has the $k$ and $w$ in the phase part. So no matter what you do to them, you cannot affect the amplitude part.

Specifically, the ratio of $y$ and $sin(kx-wt)$ is meant to cancel out the phase factor in the $y$ and leave the amplitude. By changing the frequency you only change the phase part that is canceled out anyways.


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