Is my solution right or wrong ?

[Sanosuke Sagara]

My question follow with my work to solve the question is in the attachment .

 

[vsage]

My question follow with my work to solve the question is in the attachment .

"Selected zip file is invalid or corrupted"

 

[Sanosuke Sagara]

Don't worry

don't worry, the file will not harm your computer.Just zip it and mu solution is in there.

 

[Galileo]

The file contains a windows excel document with a picture:

There a string (copper wire?) attached to an oscillilator at one end (point A). The other end of the string is hung over a pulley (point B) and a mass tied to the end of the string.
The problem is:
The length AB of the copper wire (the dot line) used in this experiment is 1.5m and its cross sectional area is 0.059mm^2
The tension in the wire is 2.0 N. If the density of the copper is 8.9 x 10^3 kg/m^3 ,show that the lowest frequency to obtain a stationary wave is about 10Hz.

Sanosuke's solution is:
Pressure P=\frac{2}{0.059\cdot 10^{-6}}=33898305 Pa
V=\sqrt{\frac{33898305}{8.9 \cdot 10^3}}=61.72 m/s

V=fl (l is wavelength)

f=\frac{V}{l}=\frac{V}{4 \cdot 1.5}=\frac{61.72}{6}=10.3 Hz \approx 10 Hz

 

[Sanosuke Sagara]

Can I conceive the system to be an air column with one end closed as the l(wavelength) is equal to 4L ?

By this way, I can obtain the answer to be at least 10Hz.

 

[siddharth]

Here treat it as a rope with both ends tied to a wall.

The Mass is given so that the tension in the rope "T=Mg" can be found
Also the mass per unit length \mu can be found using the given data

then the velocity of the wave in the string will be
v= \sqrt \frac{T}{ \mu}

Using the velocity and the fact that the wavelength will correspond to
\frac{n \lambda}{2}
The frequency can be found