Question

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end.The wire is 0.330 m long and has a mass of 9.60 g. It is fixed at both ends and oscillates in its fundamental mode.By resonance,it sets the air column in the tube into oscillation at that column’s fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

Answer 1

Answer:

Explanation:

tube:

f=v/4L = 343/(4*1.2)= 71.4583Hz tube's fundamental frequency  

wire:  

f=v/2L -> v=2Lf

v= 2*0.323*71.4583= 46.162m/s  

ρ= 0.0095/0.323= 0.02941kg/m  

v=√(T/ρ) -> T=v^2*ρ

T= 46.162^2*0.02941= 62.67[N] Tension of wire.