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This is the question sheet and my solution. If anyone finds a mistake, feel free to tell me

This is the corrected version: I corrected Example 1, because I wrote the wrong vector for . Sry for that
I also corrected example 2, I hope that future examples will be more specific.

I dont know why your fm got multiplied by (-cos;sin;0)?
At the beginning you wrote f(vector)= (-fm;0;0) and thats correct. On your local coordinate system fm(vector) should not change its angle!

could you explain me why you put in the second part of the second example T=0 in your first step? shouldnt it be only T_0=0?
thanks in advance!!

_Hofi_ hat geschrieben:I dont know why your fm got multiplied by (-cos;sin;0)?
At the beginning you wrote f(vector)= (-fm;0;0) and thats correct. On your local coordinate system fm(vector) should not change its angle!

thanks, it was late at night I guess I will correct my solution in a second

tola99 hat geschrieben:could you explain me why you put in the second part of the second example T=0 in your first step? shouldnt it be only T_0=0?
thanks in advance!!

Next to the sketch of example 2 there is drawn. I was also confused at the very beginning. I don't know if this is how it was ment to be.
If you take T(t) as unknown, you would get the same result as in the first part, but every term that has in it, you only would get instead. And then you have to correct your constants as well. I was not sure about that, so I am going to ask the Prof. on Wednesday

smefix hat geschrieben:

tola99 hat geschrieben:could you explain me why you put in the second part of the second example T=0 in your first step? shouldnt it be only T_0=0?
thanks in advance!!

Next to the sketch of example 2 there is drawn. I was also confused at the very beginning. I don't know if this is how it was ment to be.
If you take T(t) as unknown, you would get the same result as in the first part, but every term that has in it, you only would get instead. And then you have to correct your constants as well. I was not sure about that, so I am going to ask the Prof. on Wednesday

Thanks a lot, had the same thoughts like you! I calculated it both ways, for me your solution looks better, but i'm confused because we did it in the tutorial that we said T_0 is zero after the stimulus, and it would be also more logical to me, if after the stimulus T_0 ends, there is still some kind of tension which is then decreases. but you're right about the sketch so i is really confusing.....

T0 is just some "additional" tension provided by the tensile element, which counteracts the tension applied.
The graph provided in the homework tells us that someone "pulls" at the end of the muscle element and then lets go.
I guess the "boundary condition" of T(t=c) = 0 is irrelevant here, because we are looking at the displacement x instead of the tension T, and it makes sense to assume the initial tension affects the displacement of the spring, which will take a while to swing back into its initial position.

Grinsekatze hat geschrieben:T0 is just some "additional" tension provided by the tensile element, which counteracts the tension applied.
The graph provided in the homework tells us that someone "pulls" at the end of the muscle element and then lets go.
I guess the "boundary condition" of T(t=c) = 0 is irrelevant here, because we are looking at the displacement x instead of the tension T, and it makes sense to assume the initial tension affects the displacement of the spring, which will take a while to swing back into its initial position.

Today I asked the prof. about this and he said, that as long it is not explicitly written that the muscle is pulled with a force T, it should be zero. So the sum of all three forces is 0.
However, if there was a force T we would need it's exact time dependency, or it is constant in order to solve the differential equation.
Since it is bit written here it should be constant and with the prof. Advice it is constant zero.
I will correct my solution this evening.