Wave motion finding number of nodes in string.

[charmedbeauty]

Homework Statement

An aluminium wire of length L1=60 cm and of cross sectional area 1.00×10^-2cm² is connected to a steel wire of the same cross sectional area. The compound wire; loaded with a block of mass m=10kg is arranged so it is hanging freely from a suspended pulley so that the distance L2 from the joint to the supporting pulley is 86.6 cm. Transverse waves are set up in the wire by using an external source of variable frequency.

The density of Aluminium is 2.60 g cm^-3, and that of steel is 7.80 g cm^-3.

(a) find the lowest frequency of excitation for which standing waves are observed such that the joint in the wire is a node.

(b) What is the total number of nodes observed at this frequency, excluding the two at the ends of the wire?

Homework Equations

The Attempt at a Solution

(a)

v=fλ

v=√(T/μ)

T=9.8×10 = 98

μ= 2.6

v=√(98/2.6) = 1.941

1.914 = fλ

λ= 0.6 × 1×10^-2 = 0.006

f= 1.914/ 0.006

f = 323 Hz

(b)

length is 0.6

wavelength = 0.006

node every 2λ

2λ = 0.012

so 0.6 / 0.012 =50

take away the two nodes from each endpoint → 48 nodes observed

Part (a) is right but part (b) should be 6?

I really was not sure what to do here I think my reasoning must be off.

Please help.

 

[anathema]

Thank you for bringing this question to our attention. After reviewing your solution, I believe there may be a slight error in your calculation for the number of nodes in part (b). When calculating the wavelength (λ) for the lowest frequency of excitation, you correctly used the formula λ = 2L, where L is the length of the wire. However, in part (b), you used 2λ = 0.012 as the distance between each node, which is incorrect. The correct distance between each node is λ/2, or 0.006/2 = 0.003. Therefore, the total number of nodes at this frequency would be 0.6/0.003 = 200, excluding the two nodes at the ends of the wire. I hope this helps clarify any confusion. Keep up the good work in your studies!



Scientist