This preview shows page 1. Sign up to view the full content.
μ 1 = ρ 1 AL 1 / L 1 = 1 A and 11 /. A ν τρ = A similar expression holds for the wave speed in the steel section: 22 /. vA τρ = We note that the cross-sectional area and the tension are the same for the two sections. The equality of the frequencies for the two sections now leads to 11 1 2 2 2 , nL nL ρρ = where A has been canceled from both sides. The ratio of the integers is () () 33 22 2 33 1 11 0.866m 7.80 10 kg/m 2.50. 0.600m 2.60 10 kg/m L n n L ρ ρ × == = × The smallest integers that have this ratio are n 1 = 2 and n 2 = 5. The frequency is () 11 1 1 /2 /2 / . f nv L n L A τρ == The tension is provided by the hanging block and is τ = mg , where m is the mass of the block. Thus, () () () () ( ) 2 33 6 2 11 10.0kg 9.80m/s 2 324Hz. 2 2 0.600m 2.60 10 kg/m 1.00 10 m nm g f LA ρ − == = ×× (b) The standing wave pattern has two loops in the aluminum section and five loops in the steel section, or seven loops in all. There are eight nodes, counting the end points. 59. (a) The frequency of the wave is the same for both sections of the wire. The wave speed and wavelength, however, are both different in different sections. Suppose there are n 1 loops in the aluminum section of the wire. Then, L 1 = n 1 λ 1 n 1 v 1 /2 f , where λ 1
This is the end of the preview. Sign up to access the rest of the document.
This note was uploaded on 04/16/2011 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
- Spring '09