1
$\begingroup$

In my textbook, the equation for a wave is described as :

$y = A \sin{\frac{2\pi}{\lambda}(vt-x)}$

Now in this equation $x$ is variable as well as $y$. So when we will differentiate the equation by time $t$ we will also consider $x$ as a variable. But in my book, they simply differentiate the equation by considering $x$ as a constant. And so they get,

$\frac{2\pi Av}{\lambda}\cos{\frac{2\pi}{\lambda}(vt-x)}$

I don't understand why they didn't consider $x$ as a variable in the differentiation?

$\endgroup$
  • $\begingroup$ this is really a math question about the meaning partial differentiation. $\endgroup$ – ZeroTheHero Jan 19 at 23:53
2
$\begingroup$

We differentiate y with t holding x constant. This technique is called partial differentiation (we use $\partial y/\partial t$). Your textbook uses this way to measure the velocity of the vertical oscillations about a point along the string, irrespective of where it is on the string.

So, in the wave equation that you have given, we will consider $x$ to be a constant (not dependent on time - or time independent). This is because, that particular point on the string does not move anywhere else.

If you take the total derivative of y with respect to x, you are also considering the wave speed $dx/dt$, and as you say your derivative equals zero. This is because, you are not really measuring the vertical oscillations properly. In reality, as you take $\Delta y$, your wave moved by $\Delta x$ in time $\Delta t$. If you visualise it, when one peak goes by some distance, the nearby portion goes down by that same distance (while the wave has propogated further). Hence, you get $dy/dt =0$.

$\endgroup$
  • $\begingroup$ @Theoretical this is why measuring $\partial y/\partial t$ gives the speed in the vertical direction. If you have any doubt please feel free to ask. $\endgroup$ – KV18 Jan 11 at 8:38
  • $\begingroup$ So if I were to measure the velocity of the wave, what would I have to do as you pointed that $\frac{dx}{dt}=0$? $\endgroup$ – Theoretical Jan 11 at 8:48
  • $\begingroup$ @Yes. In that case, the problem is simplified to only y as a function of time , i.e. $y(t)$ $\endgroup$ – KV18 Jan 11 at 8:50
3
$\begingroup$

One of the problems with studying wave motion is that even for a one dimensional wave you have to juggle more than two variables.

You are using $y = A \sin{\frac{2\pi}{\lambda}(vt-x)}$ as an example of a wave equation with the wavelength $\lambda$ and the speed of the wave $v$ assumed to be constant.

$y$ is the displacement of a particle of the medium through which the wave is passing from its equilibrium position ie its position when there was no wave present.
$x$ is the displacement of the equilibrium position of a particle from some origin.
$t$ is the time.

Now coming back to the speed $v$ of the wave, what does that mean?
If there is a wave on water then to find the speed of the wave you might time $t$ the time it takes a crest of the water wave to travel a distance $x$.
The speed of the water wave is $\dfrac{\text{distance travelled by a crest}}{\text{time taken to travel distance}}=\dfrac xt$

You can do the same think using you wave equation.
What you need to do is to pick a value of the displacement of a particle from its equilibrium position $y$ and keep that value constant.
You could choose $y=A$ which is a crest or $y=-A$ which is a trough or any other constant value of $y$. Let's choose $y=A$ and say that at time $t_1$ the crest was at position $x_1$ and at time $t_2$ the crest was at position $x_2$ so the speed of the wave is $v = \dfrac{x_2-x_1}{t_2-t_1}$.

Using your wave equation we have $A=A \sin{\frac{2\pi}{\lambda}(vt_1-x_1)}$ and $A=A \sin{\frac{2\pi}{\lambda}(vt_2-x_2)}$ which leads to $vt_1-x_1 = vt_2-x_2 \Rightarrow v = \dfrac{x_2-x_1}{t_2-t_1}$. So $v$ is the speed of the wave.

Now this can be done another way again by keeping $y$ constant by differentiating the wave equation with respect to the time $t$.

$\dfrac {d\,\,}{dt}\left ( y = A \sin{\frac{2\pi}{\lambda}(vt-x)}\right )\Rightarrow 0= A \cos{\frac{2\pi}{\lambda}(vt-x)} \times \left( v-\dfrac{dx}{dt}\right)$

and this must be true for all times so $v-\dfrac{dx}{dt} = 0 \Rightarrow v=\dfrac{dx}{dt}$

So the speed of the wave is the rate of change of the position of the crest $\dfrac{dx}{dt}$


Confusion can arise because there is another speed which can be defined and that is $\dfrac{dy}{dt}$ which is the speed of the particles.

In this animation you will see that the speed at which a crest moves to the right which is constant is not the same as the speed at which a particle moves up and down.

enter image description here

So how does one find $\frac {dy}{dt}$?
One does that by choosing a position $x$ ie keep $x$ constant and then differentiation the equation with respect to time $t$ which is what was done in your textbook to give $\frac {dy}{dt}=\frac{2\pi Av}{\lambda}\cos{\frac{2\pi}{\lambda}(vt-x)}$.
As expected this speed is not constant unlike the speed of the wave $v$.


You might consider the wave equation $y = A \sin{\frac{2\pi}{\lambda}(vt+x)}$ with the minus sign replaced by a plus sign and show that this is a wave that moves in the negative x-direction ie $v=-\dfrac{dx}{dt}$


In terms of more advanced work using partial differentials you might write the speed of the wave as $\left(\dfrac{\partial x}{\partial t}\right )_{\rm y}$ and the speed of a particle as $\left(\dfrac{\partial y}{\partial t}\right )_{\rm x}$.

$\endgroup$
  • $\begingroup$ So can I write the partial differential of $x$ as $f\lambda$? $\endgroup$ – Theoretical Jan 12 at 9:09

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.