19. The transverse displacement u(x, t) of a string obeys...

19. The transverse displacement $u(x, t)$ of a string obeys the wave equation $\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$, where $t$ and $x$ denote time and position respectively and $c$ is a constant. Show that a possible solution is given by $u(x, t) = f(x - ct) + g(x + ct)$, where $f$ and $g$ are any functions that can be differentiated twice. Suppose that the string stretches to infinity in both directions, and satisfies the intial conditions, $u(x, 0) = 0$, $\frac{\partial u}{\partial t}(x, 0) = \frac{x}{(1 + x^2)^2}$. Show that the initial conditions can be satisfied by taking $f = -g$ for suitable $f$. Find the solution $u(x, t)$ and sketch its behaviour.

Transcript

00:01 Hello.

00:02 So it is given to us that the transverse wave's equation is y is equal to minus 0 .2, sine 16 .3t minus 180x plus 10 .455.

00:16 Now, we know that the general equation for a transverse wave is y is equal to y0 sine omega -t plus or minus x plus 5.

00:38 Okay, this is the general equation.

00:41 Here, omega is the angular frequency and k is the wave vector.

00:47 Okay, now in the first question, and 5 is the phase difference, in the first question we are asked to find the amplitude.

00:55 Okay, so the amplitude a will be equal to y 0 and when comparing with the given equation, we get the amplitude a equals 0 .2 meters.

01:09 Okay.

01:10 Now we have to find the frequency.

01:13 Okay.

01:14 So in order to find the frequency, we know that the angular frequency is omega equals 2 pi multiplied by frequency.

01:24 Okay.

01:25 So frequency will be equal to omega divided by 2 pi.

01:29 And from the equation we know that omega is equal to 16 .3.

01:37 Okay.

01:38 So, frequency f would be equal to 16 .3 divided by pi.

01:46 And on calculating, we get the answer to be equal to 2 .59 hertz, which is approximately equal to 2 .6 hertz.

02:03 Now coming to the third part of the question, we have to find the wavelength.

02:08 Okay.

02:09 Now we know that the wave vector k is equal to 2 pi divided by lambda.

02:15 Or lambda could be written to be equal to 2 pi divided by k.

02:21 Okay.

02:22 Now substituting the values in this equation, we get lambda equals 2 pi divided by k is 180 according to the equation.

02:34 To us.

02:35 Okay, so that would be equal to 0 .035 meters.

02:42 Okay.

02:43 Now, coming to the fourth part of the question, we are asked to find the angular frequency.

02:49 Now, from the general equation and comparing the equation that is given to us, the angular frequency omega could be written to be equal to 16 .3 radiance per second.

03:02 Okay.

03:04 Now, now.

03:04 Now we have to find the wave vector k and comparing the given equation and the original equation, we get the wave vector k equals 180 radiance per meter.

03:19 Okay.

03:21 Now for the next part of the equation, sorry question, we have to find the wave speed and the wave speed v could be found using the equation omega divided by the wave vector k.

03:32 Okay.

03:33 Now substituting the values in this equation, we can.

03:36 Get this is equal to 16 .3 divided by 180 and on calculating we get the answer to be equal to 0 .0905 meter per second or this is approximately equal to 0 .091 meter per second.

04:02 Okay now coming to the next part, we are asked to find the direction of propagation of the wave.

04:11 Now, if we look at the general equation that is written here, this y0 sine omega -t plus kx plus 5 is the equation for waves that are traveling in the negative x direction...

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