Ii a cord has two sections with linear densities of 010...

(II) A cord has two sections with linear densities of 0.10 $\mathrm{kg} / \mathrm{m}$ and $0.20 \mathrm{kg} / \mathrm{m},$ Fig. $34 .$ An incident wave, given by $D=(0.050 \mathrm{m}) \sin (7.5 x-12.0 t),$ where $x$ is in meters and $t$ in seconds, travels along the lighter cord. (a) What is the wavelength on the lighter section of the cord? (b) What is the tension in the cord? (c) What is the wavelength when the wave travels on the heavier section?

Transcript

00:03 Hello, so this is a wave of question and we've been giving some values.

00:06 Let's check that out.

00:08 So you're giving the linear density of the lighter cords.

00:12 I'm going to say that is m .1, that's 0 .10, kilogram per meter.

00:22 They also have the linear density of the heavier one.

00:25 I'm going to say mew 2.

00:28 That's going to be 0 .20.

00:31 Kilo kilogram meter at a wave equation d 0 .05 0 .0.

00:44 0 .5 x minus 0 t so what we're not going to do with this is we're going to compare this with the general equation the general wave equation with a sign into k x minus 12 minus omega t okay so this is the general wave equation so we're going to compare so for example if you look at the two equations you rather that k k is going to have a relationship with uh 7 .5 right because there are all the coefficients of x so we can say k is going to be equal to 7 .5 but the general formula for finding k is 2 pi over lambda so here we can get the wavelength of the lighter section of the cord.

02:12 So the wavelength of a key, right? pye over 7 .5.

02:27 It's going to be 0 .837.

02:33 So this is going to be the wavelength on the latter section of the court.

02:41 Let's talk about the next one, which is the tension.

02:48 So the speed of the wave, v, is related to the tension by t over mule.

03:04 2mu is a linear density and that is also equal to omega over k these are just general expressions we can just look up so what i can do so i'm gonna use these two right here and forget about the v and use t root t over mu equals omega k so i'm going to get rid of the root size.

03:39 I'm going to square both sides.

03:42 Let's square this side and this is going to be t over mule.

03:48 That's going to be omega squared over k squared.

03:55 So we're trying to find the tension here.

04:02 So the tension is going to be sat.

04:07 That's going to be omega squared for k squared times mule.

04:15 So now we can do a substitution...