- #1
phoebz
- 19
- 0
The figure shows an aluminum wire of length L 1 = 60.0 cm, cross-sectional area 1.00 x 10-2 cm2, and density 2600 kg/m3, joined to a steel wire of density 7.80 g/cm3 and the same cross-sectional area. The compound wire, loaded with a block of mass 10.0 kg, is arranged so that the distance L 2from the joint to the supporting pulley is 86.6 cm. Transverse waves are set up on the wire by an external source of variable frequency; a node is located at the pulley. [Hint: You may want to calculate the ratio of the number of loops in the steel compared to the aluminum. Then surmise which numbers will give you the desired lowest frequency.]
-What is the lowest frequency that generates a standing wave pattern that has the joint as one of the nodes?
*Youngs Modulus --> Aluminium: 7x10^10
Steel: 20x10^10 kg/m^3
My Attempt:
Y= (F/A)/(ΔL/L)
-The force put on both strings will be equal so I tried setting the two wire equations equal as
YAΔL/L (for aluminium) = YAΔL/L (for steel)
I was trying to find the length of the other string but I got stumped because I did have the change in length for either string.
I needed the length of each string for the equation μ=mass/length because v=√(F/μ) and frequency equals velocity/λ.
I'm not sure I even knew where I was going with this. Please help!
-What is the lowest frequency that generates a standing wave pattern that has the joint as one of the nodes?
*Youngs Modulus --> Aluminium: 7x10^10
Steel: 20x10^10 kg/m^3
My Attempt:
Y= (F/A)/(ΔL/L)
-The force put on both strings will be equal so I tried setting the two wire equations equal as
YAΔL/L (for aluminium) = YAΔL/L (for steel)
I was trying to find the length of the other string but I got stumped because I did have the change in length for either string.
I needed the length of each string for the equation μ=mass/length because v=√(F/μ) and frequency equals velocity/λ.
I'm not sure I even knew where I was going with this. Please help!