- #1

Stealth849

- 38

- 0

## Homework Statement

An aluminum wire of length l

_{1}= 60cm, cross section area 1.00 x 10

^{-2}cm

^{2}, and density 2.60g/cm

^{3}, is joined to a steel wire of density 7.8g/cm

^{3}and same cross sectional area. The compound wire loaded with a block of mass m = 10kg is arranged as shown so that the distance l

_{2}from the joint to the supporting pulley is 86.6cm. Transverse waves are set up in the wire using an external source of variable frequency. A node is located at the pulley.

a) find the lowest frequency of excitation for which a standing wave is observed to have a node at the joint.

b) How many nodes are observed at this frequency?

## Homework Equations

f = v/λ = nv/2l

ρ = m/V

v = sqrt(F

_{T}/μ)

## The Attempt at a Solution

Not too sure where to begin on this one. When it asks for the lowest frequency of excitation, I expect that means the second harmonic, because that would leave a node at the joint with the lowest frequency.

So knowing the densities of the wires, I can get v where

F

_{T}= μv

^{2}

F

_{T}= mg = 98N

μ = m/l = ρV/l = ρA where A represents the cross sectional area.

so μ

_{1}= 2600*0.0001 = 0.26

μ

_{2}= 7800*0.0001 = 0.78

v

_{1}= sqrt(98/0.26) = 19.4m/s

v

_{2}= sqrt(98/0.78) = 11.2m/s

so f = nv/2l

where n = 2 for the second harmonic

f

_{1}= 2*0.26/(2*0.6) = 0.43Hz

f

_{2}= 2*0.78/(2*0.866) = 0.90Hz

No idea if I'm approaching this correctly, due to the idea of reflection of waves at the joint.. Say a wave does reflect at the joint and a portion of it is sent back to the source, there would be a lot more waves thus being a much higher harmonic - not the second.

Any and all help appreciated. Thanks!

#### Attachments

Last edited: