One end of a horizontal string of linear density 6.6*10^-4 kg/m is attached to a small amplitude mechanical 120-Hz oscillator. The string passes over a pulley a distance of l=1.50 m away, and weights are hung from this end. What mass m must be hung from this end of the string to produce (a) one loop, (b) two loops, and (c) five loops of a standing wave? Assume the string at the oscillator is a node, which is nearly true.
µ = Γ = 6.6 * 10^-4 kg/m
f = 120 Hz
L = 1.50 m
g = 9.8 m/s^2
F_t = mg
v = √(F_t*L/m)
v = Γf
Γ = 2L/n where n = 1 for first harmonic, n = 2 for second harmonic, etc.
The Attempt at a Solution
I'm really not sure where to go with this one, so I'm just going to solve for what I can.
First, I'll solve for velocity.
v = Γf = (0.00066 kg/m)(120 Hz) = 0.0792 m/s
Now I'm going to set the velocity formulas equal to each other to try to solve for m.
√(F_t*L/m) = Γf
√(mg*L/m) = 0.0792
√(g*L) = 0.0792
√(9.8*1.5) = 0.0792
Oh wait, mass cancelled itself out >.<
Help is much appreciated!