How Does Tension Affect the Frequency of Standing Waves?

[tomasblender]

Homework Statement

In the arrangement shown in the figure, a mass can be hung from a string (with a linear mass density of μ=0.00182 kg/m) that passes over a light pulley. The string is connected to a vibrator (of constant frequency f), and the length of the string between point P and the pulley is L=1.95 m. When the mass m is either 16.5 kg or 22.5 kg, standing waves are observed; however no standing waves are observed with any mass between these values.

http://capa.physics.mcmaster.ca/figures/sb/Graph18/sb-pic1825.png

What is the frequency of the vibrator? (Hint: The greater the tension in the string the smaller the number of nodes in the standing wave.)

What is the largest mass for which standing waves could be observed?

Homework Equations

f(m) = mf1
f = (T/u)^.5/2L

The Attempt at a Solution

(16.5*9.8/0.00182)^.5/(2*1.95) = 79.249 Hz = (m + 1)f
(22.5*9.8/0.00182)^.5/(2*1.95) = 89.249 Hz = (m) f

Therefore

79.249 - 89.249 = mf + f - mf
10.000 Hz
Which is the incorrect answer.

would be

(mg/u)^.5/2L = Answer for part A
("a"^2)*(2L^2)*u/g = m

I cannot test if this is the answer without having the answer for the first section. Anyways I am dry out of ideas for part A, any help?

 

[ideasrule]

f(m) = mf1
f = (T/u)^.5/2L

The Attempt at a Solution

(16.5*9.8/0.00182)^.5/(2*1.95) = 79.249 Hz = (m + 1)f
(22.5*9.8/0.00182)^.5/(2*1.95) = 89.249 Hz = (m) f

You assumed that the fundamental frequency is the same in both scenarios. That's not true, since the tensions are different, resulting in different fundamental frequencies.

Try writing out the wavelength for both cases as a function of m, u, and f. Also write out the wavelength in terms of L and n1/n2 (representing the n1/n2th harmonic). Then you'll be able to find n1 and n2, which will tell you f.