Vibrating aluminium string

[kudoushinichi88]

Homework Statement

An aluminium block of m is hung from a steel wire of length L. The fundamental
frequency for transverse standing waves on the wire is 300 Hz. The block
is then immersed in water so that half of its volume is submerged. What is the
new fundamental frequency? (You may assume that the mass of the wire is small
compared to the mass of the block and the change in length of the wire under
different loads is negligible.)

Homework Equations

Speed of wave on a string,

[tex]v=\sqrt{\frac{T}{\mu}}[/tex]

Buoyancy force

[tex]F=\rho g V[/tex]

The Attempt at a Solution

[tex]\frac{fL}{2}=\sqrt{\frac{T}{\mu}}[/tex]

when suspended in air,

[tex]150L=\sqrt{\frac{mg}{\mu}}[/tex]

When half of its volume immersed in water,

[tex]\frac{fL}{2}=\sqrt{\frac{mg-\frac{\rho_{water}gV}{2}}{\mu}}=\sqrt{\frac{mg-\frac{\rho_{water}mg}{2\rho_{Al}}}{\mu}}[/tex]

The answer I got is

[tex]f=300\sqrt{1-\frac{\rho_{water}}{2\rho_{Al}}[/tex]

Subbing in values gives me a value of 270Hz...

are my steps correct?

 

[alphysicist]

Hi kudoshinichi,

Homework Statement

An aluminium block of m is hung from a steel wire of length L. The fundamental
frequency for transverse standing waves on the wire is 300 Hz. The block
is then immersed in water so that half of its volume is submerged. What is the
new fundamental frequency? (You may assume that the mass of the wire is small
compared to the mass of the block and the change in length of the wire under
different loads is negligible.)

Homework Equations

Speed of wave on a string,

[tex]v=\sqrt{\frac{T}{\mu}}[/tex]

Buoyancy force

[tex]F=\rho g V[/tex]

The Attempt at a Solution

[tex]\frac{fL}{2}=\sqrt{\frac{T}{\mu}}[/tex]

I think your final expression at the end of your post is correct. But this expression is not quite right; the fundamental wavelength is 2L, not L/2. However, in this problem the wavelength will cancel out.

when suspended in air,

[tex]150L=\sqrt{\frac{mg}{\mu}}[/tex]

When half of its volume immersed in water,

[tex]\frac{fL}{2}=\sqrt{\frac{mg-\frac{\rho_{water}gV}{2}}{\mu}}=\sqrt{\frac{mg-\frac{\rho_{water}mg}{2\rho_{Al}}}{\mu}}[/tex]

The answer I got is

[tex]f=300\sqrt{1-\frac{\rho_{water}}{2\rho_{Al}}[/tex]

Subbing in values gives me a value of 270Hz...

are my steps correct?

 

[Delphi51]

Again, every line checks. Very clever of you to eliminate the unknown V that way.
Okay, I see the error Alphysicist points out. Thank you.

 

[kudoushinichi88]

Oh! -_-"

Carelessness... Well, I guess I need to sleep. It's 4.30am here...

Thank you for your insight! I appreciate it a lot!

 

[rwisz]

Yes, your steps are correct. Your final answer of 270 Hz is the correct new fundamental frequency for the vibrating aluminium string when half of its volume is submerged in water. This is because the buoyancy force from the water reduces the effective mass of the string, resulting in a higher frequency. Good job!