0:00
Hi there.
00:01
So for this problem, we have an aluminum wire of length that we're going to call the length 1.
00:08
And that length is equal to 60 centimeters and a conceptual area.
00:14
We're going to call this conceptual area 1.
00:17
And that is equal to 1 times 10 to the minus 2 centimeters square.
00:27
And a density, we're going to call the density 1, and that density is equal to 2 .60 grams per centimeter, or cubic centimeter, and is joined to a steel wire with a density, the we're going to call the density 2, that is 7 .80 grams per cubic centimeter.
00:55
And the same conceptional area.
00:59
So the conceptual area 2 is the same as the corceptional area 1.
01:04
We're going to call just the corceptional area a.
01:08
Now, the compound wire loaded with a block of mass, m, that is also given its value, is 10 kilograms, is a range so that the distance l2 from the joint to the supporting pulley is 80 to 86 .6 centimeters.
01:28
So the distance l2 is also given.
01:37
Transerve waves are set up on the wire by an external source of variable frequency.
01:46
A node is located at the pulley.
01:48
So for part a of this problem, we need to find the lowest frequency that generates a standing wave having the joint as one of the nodes.
01:57
So that's what we need to find for part a of this problem.
02:01
So with that said, we know that the frequency of the wave is the same for both sections of the wire.
02:14
The wave speed and wavelength, however, are both different in different sections.
02:21
And so we start by supposing that there are n1 loops in the aluminum section of the wire.
02:29
Then the distance l1 can be written as the number n1.
02:36
This is again the number of loops in the aluminum section times the wavelength in that section in this divided by 2.
02:46
Now we can express this in terms of the frequency.
02:50
So that will be n1 times the speed 1.
02:54
And this divided by 2 times the frequency that we're going to call just simply as f because it is the same for both sections of the wire.
03:05
Now, in this consideration, we have substitute, of course, the relation that it sits between the wavelength and the speed with the frequency, just like this.
03:19
Now, since we want to solve for the frequency from this previous expression, we obtained that that is the number n1 times the speed 1, and this divided by two times the length 1.
03:36
So a similar expression holds for the stillception, so something similar, but for the still exception, we will have that that is just simply n2 times the speed 2, and this divided by two times the length.
03:58
And since the frequency is the same for the two sections, we can equal these two expressions.
04:06
So we will have n1 times the speed 1, and this divided by the length 1, we can cancel the 2's, and then this is n2 times the speed divided by the length 2.
04:21
Now, the width speed in the aluminum section is given by speed 1, is the square root of the tension, and this divided the, by the linear mass density.
04:52
That we're gonna call musu 1 for the aluminum, for the aluminum wire.
05:00
Now, the mass of the aluminum in the wire is given by, it's called just the mass m1, mass m1 is just simply the density of the aluminum times the cross -ceptional area, which we are given in that is a, times the length of that wire.
05:21
L1.
05:23
So we can write then that the linear density which is just simply the mass m1 divided by the length.
05:42
So we will obtain that that is to simply the density times the corosceptional area.
05:50
So we will have that then the speed can be written as the square root of the tension in this cable, and this divided by the expression that we have for the linear density mass, which is just simply the period between the density and the perceptional area.
06:15
Now, a similar expression, of course, holds for the wave speed in the steel section.
06:21
Now, but in that case, that speed is just simply the square root of the tension, divided by the density of the steel wire that we call the density two times the perceptional area a...