- #1

gmmstr827

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## Homework Statement

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

## Homework Equations

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!

Thank you!