Express your answer in terms of the variables v_max and a_max.
Find the period of this motion.
Express your answer in terms of the variables v_max and a_max.
I don't even know where to start. Thanks for the help!An object executing simple harmonic motion has a maximum speed vmax and a maximum acceleration amax?To answer the second part of your question we again turn to the equations we have for v_max and a_max; however, this time instead of using the v_max=A x W this time were going to substitute another equation in for one of these variable.
Because we are looking for T (period) we need to find an equation that uses this variable, and it just so happens we have one where W= (2蟺/T). Thus we start this problem now by writing our two equations:
v_max= (2蟺/T) x A and a_max= (2蟺/T)^2 x A
Next, we want to solve each of these equations for T, thus giving us:
v_max:
T= (2蟺 x A)/(v_max)
and,
a_max:
T= (2蟺)/square root[(a_max)/A]
Now, if you can follow what has happened thus far, we set these two T (periods) equal to each other and solve them to equal A, which we find to be: A= (v_max)^2/(a_max)
We take this equation for the value of A and substitute it back into one of our original T= equations, thus in the end yielding us with our result:
T= [2蟺 x (v_max)^2/(a_max)]/(v_max)
Hope this helps, many apologies for my inabilities to create mathematical symbols on my computer but I hope this answer serves you wellAn object executing simple harmonic motion has a maximum speed vmax and a maximum acceleration amax?well this is how i did it.
(w = omega)
a_max = Aw^2
v_max = Aw
so then i solved for A in each of those to get A= a_max/w^2 and A= v_max/w so then we have 2 equations for A we can set equal to each other and then just solve for omega so you eventually get w = (a_max) / (v_max).
so then you can plug that w back into one of the original equations for A
A= v_max / w
so
A = v_max / (a_max/v_max) so then just simplify so your final answer should be
A = (v_max)^2 / (a_max)
No comments:
Post a Comment