Problem
We were posed the problem to find the Linear Density of a string using only an oscillator, pulley and standing weights. We were trying to find µ by using the knowledge that has been acquired over our waves unit.
Materials
For this experiment we were only given an oscillator, a meter stick, a pulley, a string and standing weights.
Procedure
We started by determining the equation for µ and how it relates to linear density.
First we started out with the equation v = λƒ and then solved for µ using that.
v = λƒ
v = √(T/µ)
λƒ = √(T/µ)
(λƒ)^2 = T/µ
µ = T/(λƒ)
µ = mg/(λƒ)^2
Then we had to come up with a procedure to match this and find the missing variables. We started with finding the tension of the spring which is equal to mass multiplied by gravity. The mass of the string is negligible so the mass of the system is just the hanging mass. The hanging mass 200 grams or .2 Kilograms. Therefore, the tension of the spring is 2 Newtons since the force of gravity is 10 Newtons. Now that we have that we set up the experiment so that we can find the other variables. The picture above shows the lab setup.
First we started out with the equation v = λƒ and then solved for µ using that.
v = λƒ
v = √(T/µ)
λƒ = √(T/µ)
(λƒ)^2 = T/µ
µ = T/(λƒ)
µ = mg/(λƒ)^2
Then we had to come up with a procedure to match this and find the missing variables. We started with finding the tension of the spring which is equal to mass multiplied by gravity. The mass of the string is negligible so the mass of the system is just the hanging mass. The hanging mass 200 grams or .2 Kilograms. Therefore, the tension of the spring is 2 Newtons since the force of gravity is 10 Newtons. Now that we have that we set up the experiment so that we can find the other variables. The picture above shows the lab setup.
In this the spring is put on a pulley at one end and on the other attached to an oscillator that will determine the frequency. We first put the weights on the end of the spring that had gone through the pulley and made sure that the string was taught. Once this was the case we started to adjust the frequency until the wave that appeared looked like the first harmonic. We found that the frequency needed to achieve this harmonic was 11 hz.
Once we achieved this first harmonic, we had to measure the length of the wave that had form. For the first harmonic, the equation is 2 times the length of the string so after converting our measurement to meters and doubling it we concluded that the wavelength was 2.1 meters. Now that we know all of the variables needed to solve for the linear density we can just plug them into our previous equation.
µ = mg/(λƒ)^2
µ = 2N/(2.1*11)^2
µ = 0.032
This shows that the linear density of the string was 0.0037 kg/s
µ = mg/(λƒ)^2
µ = 2N/(2.1*11)^2
µ = 0.032
This shows that the linear density of the string was 0.0037 kg/s
To confirm that this number was accurate we decided to do two more trials in which we would change the frequency and therefore the wavelength of the wave. To do so we started to look for the second and third harmonics. We found that their frequency was 22 Hz and 33 Hz respectfully. This would in turn change the wavelength since the medium that the wave was traveling through was still constant and therefore the velocity of the wave stays the same.
This is the second harmonic for the wave that we had created earlier. It has a frequency of 22 Hz and has a wavelength of 1.05 meters since the wavelength of the second harmonic is equal to the length of the string. The tension that is applied is constant from the last experiment so we now have all of the variables needed to find the linear density.
µ = mg/(λƒ)^2
µ = 2N/(1.05*22)^2
µ = 0.0037
This shows that the linear density of the string is 0.0037 kg/s
µ = mg/(λƒ)^2
µ = 2N/(1.05*22)^2
µ = 0.0037
This shows that the linear density of the string is 0.0037 kg/s
This is the third harmonic for the wave that we had created earlier. It has a frequency of 33 Hz and has a wavelength of 0.70 meters since the wavelength of the third harmonic is equal to the length of the string. The tension that is applied is constant from the last experiment so we now have all of the variables needed to find the linear density.
µ = mg/(λƒ)^2
µ = 2N/(0.70*33)^2
µ = 0.0037
The linear density of the string is 0.0037 kg/s
µ = mg/(λƒ)^2
µ = 2N/(0.70*33)^2
µ = 0.0037
The linear density of the string is 0.0037 kg/s
Conclusion
In conclusion, the linear density of the string is 0.0037 kg/s. This was proved by three different trials using the 1st, 2nd and 3rd harmonics which all showed that they had the same linear density which gives a low chance of error due to the repeated trials. Therefore, I am confident that the linear density is 0.0037 kg/s.