# Standing waves and oscillation

## 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!

zay47olali

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rl.bhat
Homework Helper
Here
v = (Mg*L/m)^1/2, where M is mass hung from the other end of the string, and m is the mass of the string.

Here
v = (Mg*L/m)^1/2, where M is mass hung from the other end of the string, and m is the mass of the string.
Okay, so I would plug everything in and solve for M?

0.0792 = (9.8M*1.5/0.00066)^1/2
M = 2.82 * 10^-7 = 0.000000282 kg

This seems like a very small mass. Is this how it's solved?

rl.bhat
Homework Helper

Do I use µ = m/L >>> µL=m where µ = 6.6*10^-4 kg/m and L = 1.50 m
so the mass of the string = 0.00099 kg
then
Γ = 2L/n where n=1 and L = 1.50 m so
Γ = 3
So 3*120 = 360 m/s

v = √(Mg*L/m)
360 = √(M*9.8*1.5/(9.9 * 10^-4))
m = 8.73 kg?

So for part b, I would just use 2 instead of 1 for n; and part c I would use 5. Yes?

b) v = 180 m/s; m = 2.18 kg
c) v = 48 m/s; m = 0.155 kg

Last edited:
rl.bhat
Homework Helper
Seems good.