course phy232
15.74Holding up stress and strain
A string or rope will break if it is placed under too much tensile stress. Thicker rope can withstand more tension without breaking because the thicker the rope, the greater the cross sectional area and the smaller the stress. One type of steel has a density of 7800 kg/m^3 and will break if the tensile stress exceeds 7.0 * 10^8 N/m^2. You want to make a guitar string from this 4.0g type of steel. In use, the string must be able to withstand a tension of 900 N without breaking. Your job is the following: Determine the maximum length and minimum radius the string can have:
note that the above could have been abbreviated to something like 'find the max length and min radius for a 4 g string able to withstand tension 900 N. rho = 7800 kg/m^2, max stress 7*10^8 N /m^2'. The full wording looks great, of course, but I you can reduce the typing load.
- Taking the tensile stress= F/A we can use
(7.0 * 10^8 N/m^2)= 900 N / A and find the area, which comes out to be 1.2*10^-6 m^2
Taking the mass of the guitar string and dividing it by its density= (4* 10^-3 kg) / 7800 kg/m^3 we get its volume which is 5.2 * 10^-7 m^3
We know the volume, and we know that area * length gives volume, so we can take the volume and divide by the area (I'm guessing they are assuming this is a cylindrical shaped string so its area is equivalent to pi * r^2
The length will be Volume/ Area = (5.2* 10 ^ -7 m^3) / (1.2* 10 ^ -6 m^2) we get a length of .43m
To find the radius the string can have we can take the area and divide it by pi and sqrt it .
sqrt ((1.2* 10^ -6 m^2 ) / pi) = r
very good solution; the process is fine and the results are reasonable. I didn't check the numbers precisely but if there's an error it's not a big one.
- Determine the highest possible frequency of standing waves on this string, if the entire length of the string is free to vibrate
I just took the fact that v= sqrt(F/ 'mu) and got the velocity v= sqrt( 900N/ (4.00*10^-3 kg/ .43m) and got a velocity of 311 m/s
then I took v= 'lamba * frequency and got frequency = 311m/s / .43m
(not too sure on this part - hopefully the first part is right, it feels like I did it correctly ) "
The fundamental frequency has a wavelength double the length of the string, so you would divide by .86 m. You get around 380 s^-1, meaning 380 Hz or 380 cycles / second.