Simple Harmonic Motion Question When finding the displacement-time function given the velocity-displacement function using integration, why do we assume v is positive? EDIT: Question is in comments.
v^2=n^2(A^2-x^2) v=n(A^2-x^2)^(1/2) Why isn't v =+-n(A^2-x^2)^(1/2)
\[v^{2}=n ^{2}(A ^{2}-x ^{2})\] \[v=n \sqrt{a^{2}-x ^{2}}\]
If you need even the direction of velocity, then choose the sign depending on whether x>0 or x<0.
But if you're just trying to find the displacement-time function, wouldn't your final answer be +- [the function]? But that wouldn't really make sense
Yeah, if we consider the +ve sign and integrate, we get a trigonometric function which automatically takes care of sign.
But depending on the sign, after we integrate, it will change to either arcsin or arccos...How do we figure out which one it is
There should be some information of position and time, which can help us finding the appropriate function.
A particle is moving with simple harmonic motion of period pi seconds and maximum velocity 8m/s. If the particle started from rest at x=a, find a.
I got up to the step v^2=64-4x^2 but what do I do when I squareroot both sides? Do I put a +- or assume positive (like my textbook does). But why do you assume positive?
Why don't you post it in physics? I will reply there..
In Australia, motion is in mathematics for some reason haha :)
That is good. :) You can solve this way, Use, \[T = 2\pi /\omega \] where T-time period, w- angular velocity. Now, use the expression \[v = A \omega \cos wt \] For maximum velocity. Got it?
hmm but the book says to integrate to get an inverse trig function, and then rearrange the equation so its like x=[trig function]. Then you substitute t=0 in to find what a is
Oh, then consider the positive sign. the sin or cos will take care of the sign. anyway, the w above was angular frequency. sorry.
okay! I think I kind of get it... Is it like you can use positive or negative, but whichever you use, the trig function will take care of it for you?
in the question above though, you have to take v as negative though
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