experimental-physicsforcesnewtonian-mechanicsstring
below is an experiment set up to test the effect tension has on the frequency of various guitar strings. It's hard to see the string in the picture, but it's attached to a weight and I was wondering how you could calculate the tension on the string? What equation/formula would you use? We know the length, mass of the string.
The speed of sound according to this chart would change as much as 330.4/358.0 = 8% . The air behaves closely to an ideal gas, therefore this kind of change would hardly change the pitch, let alone be noticeable to a human ear.
However the change in temperature would effect the stiffness and length of the string dramatically, this depends on what material you are using. Look at the difference of length vs temperature.
Since you are using weights, your stress is not going to change, and the tension would remain the same.
Try using this equation:
But this is not the full picture, since you are actually using an open string with dampening. Though with those speeds it should be approximated to a closed string.
You can try using: $$f_open = \frac{\sqrt{\frac{T}{m/2}}}{2L} \cdot constant$$
And tune the constant to fit your current parameter space.
Source with calculator (The calculator though would not work in your case because one of your ends is open).
That is a really good question that usually books fail to address in an intuitive way. Let me give you a step by step intuitive explanation by looking at the Atwood's machine:
Start with the balance of forces around the cable:
$$𝑭_𝟐−𝑭_𝟏=𝒎𝒂$$
F2 is greater than F1 since the cube has more mass than the sphere. Now we use the common assumption that textbooks use, that is, assume that the cable's mass is insignificant: $$m = 0$$
So the first equation would yield:
$$𝑭_𝟐−𝑭_𝟏=0(=) 𝑭_𝟐 = 𝑭_𝟏 $$
What is strange is that we are applying different forces at the cable's ends, however, the cable "feels" two equal forces. But we clearly know that F2 and F1 aren't the same. So what is this force that the cable "feels" at its extremities? Let's call it tension, T:
Since the cable is "feeling" two equal forces, T, at its extremities, then it will react against the bodies it is attached to, with also a value of T. That's why in the Atwood's machine from the first figure, a tension T is applied against each body. Remember this is only true with the initial assumption (m=0), and a more realistic approach is mentioned in danimal's answer.
Of course, you now can calculate what is the value of T by analyzing the FBD of the sphere and the cube. I'll leave that up to you.
Clue: assume that both bodies have the same acceleration: if the position of the cube goes downwards 1 unit, then the position of the sphere goes upwards also 1 unit (thanks to the cable which has a constant length). Since the displacements (absolute value), $r(t)$, of the bodies, are the same over time, then their derivatives are also the same, meaning that $a_1 = a_2$, where $a_i$ is the resultant acceleration on body i.
Best Answer