# Problem in finding the velocity of a wave

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?

• this is really a math question about the meaning partial differentiation. – 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 $$dy/dt =0$$.

• @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. – KV18 Jan 11 at 8:38
• So if I were to measure the velocity of the wave, what would I have to do as you pointed that $\frac{dx}{dt}=0$? – Theoretical Jan 11 at 8:48
• @Yes. In that case, the problem is simplified to only y as a function of time , i.e. $y(t)$ – 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.

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

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