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?

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

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 $dy/dt =0$.

  • $\begingroup$ @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

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 $\left(\dfrac{\partial y}{\partial t}\right )_{\rm x}$.

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

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