Question

A wire is composed of aluminum with length ℓ1=0.300m and mass per unit length μ1=1.48g/m joined to a steel section with length ℓ2=0.800m and mass per unit length μ2=0.632g/m. This composite wire is fixed at both ends and held at a uniform tension of 124 N .

Find the lowest frequency standing wave that can exist on this wire, assuming there is a node at the joint between aluminum and steel？

How many nodes (including the two at the ends) does this standing wave possess?

Answer 1

so you have to have a frequency that sets up a standing wave in both wires, with a node at each end of each wire.

Your equation is correct. With a tension in Newtons, you'll need mu to be in kg/m, so that's μ1=1.48g/m and μ2=0.632g/m.

That gives v1 = (124/0.00148)^1/2=289.45 m/sec

v2 =(124/0.000632)^1/2 = 442.94 m/sec

v1 * n1/(2L1) = v2 * n2/(2L2)

n2/n1 = (v1/v2)*(L2/L1) = (289.45/442.94)(0.800/0.300) = 1.7426 it's nearly equal to 7/4 .
289.45*4/(2*0.300) = 1930 Hz
442.94*7/(2*0.800) = 1937 Hz

so, minimum frequency=1930 Hz

no. of nodes=13