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## Main Question or Discussion Point

I am just starting with differential equations, and...I need a lot of reading, but this question should be simple for you.

Acceleration at a spring is analogous to the position (displacement) x. But the definition of acceleration is the second derivative of the function: (distance x in relation to time t), that is, the second derivative of the function f(t) where f(t)=x. So I guess the known formula is f''(t)=kx. So I want to find acceleration in relation to time (and not in relation to x) in the case of the spring. So I GUESS I have to solve this differential equation:

f’’(t)=kf(t)

(where k is the constant of the spring. Mass gets involved, but let's ignore it for the moment even if it wrong to ignore it.) And I guess that I still have to keep f''(t) in the solution (since I do want acceleration in the equation=solution), and have to get rid of f(t) only and replace it with ... what is equal to, having t as the variable instead of f(t). Lol, confused, but: I just want to find f(t)=? and then I will know f''(t) too.

LOL, the answer is known and it's f''(t)=cos(t) or something like that. And I remembered that it is known, when I thought that the cos<->sin functions are the only ones where the f’’(t)=kf(t) could be valid. (I semi-solved the diferential equation almost by intuition). They produced this proof out of Work (!!!), whereas I just prooved that this is always the case when acceleration is analogous to displacement, and exactly because of f''(t)=cf(t), and not because of Work.

Am I correct? I still don't know how to solve that though.

Acceleration at a spring is analogous to the position (displacement) x. But the definition of acceleration is the second derivative of the function: (distance x in relation to time t), that is, the second derivative of the function f(t) where f(t)=x. So I guess the known formula is f''(t)=kx. So I want to find acceleration in relation to time (and not in relation to x) in the case of the spring. So I GUESS I have to solve this differential equation:

f’’(t)=kf(t)

(where k is the constant of the spring. Mass gets involved, but let's ignore it for the moment even if it wrong to ignore it.) And I guess that I still have to keep f''(t) in the solution (since I do want acceleration in the equation=solution), and have to get rid of f(t) only and replace it with ... what is equal to, having t as the variable instead of f(t). Lol, confused, but: I just want to find f(t)=? and then I will know f''(t) too.

LOL, the answer is known and it's f''(t)=cos(t) or something like that. And I remembered that it is known, when I thought that the cos<->sin functions are the only ones where the f’’(t)=kf(t) could be valid. (I semi-solved the diferential equation almost by intuition). They produced this proof out of Work (!!!), whereas I just prooved that this is always the case when acceleration is analogous to displacement, and exactly because of f''(t)=cf(t), and not because of Work.

Am I correct? I still don't know how to solve that though.

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