In Fig. 16-43, an aluminum wire, of length L1=60.0 cm, cross-sectional area 1.00 ×10^-2 cm

In Fig. 16-43, an aluminum wire, of length L1=60.0 cm,...

In Fig. $16-43$, an aluminum wire, of length $\quad L_{1}=60.0 \quad \mathrm{~cm}$, cross-sectional area $1.00$ $\times 10^{-2} \mathrm{~cm}^{2}$, and density $2.60 \mathrm{~g} / \mathrm{cm}^{3}$, is joined to a steel wire, of density $7.80 \mathrm{~g} / \mathrm{cm}^{3}$ and the same cross-sectional area. The compound wire, loaded with a block of mass $m=10.0 \mathrm{~kg}$, is arranged so that the distance $L_{2}$ from the joint to the supporting pulley is $86.6 \mathrm{~cm}$. Transverse waves are set up on the wire by an external source of variable frequency; a node is located at the pulley. (a) Find the lowest frequency that generates a standing wave having the joint as one of the nodes. (b) How many nodes are observed at this frequency?

In Fig. $16-43$, an aluminum wire, of length $\quad L_{1}=60.0 \quad \mathrm{~cm}$,
cross-sectional area $1.00$ $\times 10^{-2} \mathrm{~cm}^{2}$, and density
$2.60 \mathrm{~g} / \mathrm{cm}^{3}$, is joined to a
steel wire, of density $7.80 \mathrm{~g} / \mathrm{cm}^{3}$ and the same
cross-sectional area. The compound wire, loaded with a block of mass $m=10.0 \mathrm{~kg}$, is arranged so that the distance $L_{2}$ from the joint to the supporting pulley is $86.6 \mathrm{~cm}$. Transverse waves are set up on the wire by an external source of variable frequency; a node is located at the pulley.
(a) Find the lowest frequency that generates a standing wave having the joint as one of the nodes. (b) How many nodes are observed at this frequency?

Key Concepts

Standing Waves

Standing waves are patterns formed when two waves of identical frequency and amplitude travel in opposite directions and interfere, resulting in fixed points of zero displacement called nodes and points of maximum displacement known as antinodes. This concept is crucial for understanding the resonance phenomena in vibrational systems.

Wave Speed and Tension

The speed of a wave traveling along a string is determined by the tension in the string and its linear mass density, according to the formula v = ?(T/?). This relationship between tension and mass density is essential for determining the frequencies and wavelengths of the standing wave modes on the string.

Boundary Conditions

Boundary conditions are the constraints imposed on a vibrating medium at its endpoints or at interfaces within the medium. They dictate whether a point behaves as a node or an antinode, which in turn determines the allowable wavelengths and frequencies for standing waves in the system.

Interface Conditions in Compound Media

In systems composed of multiple materials, such as a compound wire, the continuity of displacement and the matching of tension at the interface between different materials are important. These conditions ensure proper wave propagation and influence the standing wave patterns across the composite system.

Transcript

00:01 Hi there, so for this problem we are told in an anumelon wire of length l1 equals to 60 centimeters in cross -sectional area that is equal to 1 times 10 to the minus 2 centimeters to the square in density that is equal to 2 2 .6 grams per cubic centimeter is joined to a steel wire, we're going to call this density 1.

00:51 The density 2 is equal to 7 .8 grams per cubic centimeter in the same receptional area.

01:05 So the component wire loaded with a block of mass m that is equal to 10 kilograms and is a range so that the distance l2 from the joint to the supporting pulling, so the distance l2 is equal to 806 .6 centimeters.

01:41 So, transverse waves are set up on the wire by an external source of variable frequency.

01:50 So a node is located at the pulley.

01:54 So for part a of this problem, we need to find the lowest frequency that generates a standing wave having the join as one of the nodes.

02:10 So we know that the frequency of the wave is the same for the wave.

02:15 Both sections of the wire.

02:18 So the wave speed and wavelength, however, are both different in different sections.

02:26 So we suppose that there are any n1 loops in the aluminum section of the wire, then we will have that the length l1 is equal to n1 nodes times the wavelength 1 divided by 2.

02:47 And we now substitute the definition of the wavelength that we know in terms of the speed is the speed divided by the frequency.

03:03 Where lambda 1 is the wavelength and b1 is the wave speed in that section.

03:11 Now, in this consideration, we have submitted that lambda is equal to the speed over the frequency.

03:21 Now, if we solve for the frequency in this expression, we found that that is the number of notes n1, the speed b1 divided by 2 times the length l1.

03:38 So a similar expression holds for the steel exception for the steel.

03:46 We have that the frequency changes to the number of notes and two, the speed two, and two times the length two.

04:04 Since the frequency is the same for the two sections, we can equal these two expressions.

04:10 So we will find that n1 times the speed 1 divided by the length 1 is equal to the number of note.

04:23 2 times the speed 2 divided by the length 2.

04:28 Now the wave speed in the aluminum section, we know that that is given by tally divided by or the tension divided by the linear mass density of the aluminum wire.

04:53 The mass of the aluminum in the wire is also m1 is given by the density, the coroceptional area, and the length l1.

05:19 So from there, we find that the linear mass density is equal to the density 1 of the aluminum times the coreceptional area times its length divided by the length l1.

05:40 So we can simplify this as the density times the contraceptional area.

05:49 So with that set, we're going to have that.

05:53 The speed 1 is equal to the square root of the tension divided by the density times the contraceptional area...

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