Exercise combining Wave and Hydrostatic: Pulley and Submerged Ball

[ricmacas]

Homework Statement

One end of a horizontal string is attached to the wall, and the other end passes over a pulley. A sphere of an unknown material hangs on the end of the string. The string is vibrating with a frequency of 392 cycles per second. A container of water is raised under the sphere so that the sphere is completely submerged. In this configuration, the string vibrates with a frequency of 343 cycles per second.
What is the density of the sphere?

Homework Equations

v=\sqrt{T/\rho}
v=\lambda * f
W=\rho*V*g
F=W-B=\rho*V*g - \rho_{H20}V_{i}g (assuming W>B)

The Attempt at a Solution

Wave Perspective:
T_{1} = \rho * \lambda_{1}^{2} * f_{1}^{2}
T_{2} = \rho * \lambda_{2}^{2} * f_{2}^{2}
Mechanical Perspective:
T_{1} = W = \rho*Vg
T_{2} = W - B = \rho*Vg - \rho_{H20}V_{i}g
Thus:
T_{1} = W then, assuming displaced volume = volume of the sphere,
V_{i}g = \lambda_{1}^{2} * f_{1}^{2}
T_{2} = W - B = \rho*\lambda_{1}^{2} * f_{1}^{2} - \rho_{H20}*\lambda_{1}^{2} * f_{1}^{2}

Now, I have arrived to the answer in the solutions (it's the right answer, but negative, I can't explain that either), but I had to assume that \lambda_{1}=\lambda_{2}. Can someone explain me , if frequencies are different, why should I assume the wavelenght is the same before and after submerging the ball?

For those who want the answer:
4267 Kg m-3, which i obtained from T_{2}=T_{2} \Leftrightarrow \rho * \lambda^{2} * f_{2}^{2} = \rho\lambda^{2} * f_{1}^{2} - \rho_{H20}\lambda^{2} * f_{1}^{2}
thus \rho = (-\rho_{H20}*f_{1}^{2})/ (f_{2}^{2} - f_{1}^{2})
(The formula gives me a negative result though).

 

[TSny]

Hello, ricmacas. Welcome to PF!

The equation for the speed of a wave on a string is $v = \sqrt{T/\mu}$ where $\mu$ is the mass per unit length of the string. So, you'll need to rethink the mathematical steps.

In order to see why the wavelength should be the same, keep in mind that the standing waves must have a node at the wall and a node at the pulley.