Asked by Dan
A particle of mass m = 0.40 kg oscillates on the end of a spring with an amplitude of 0.30 m. At t = 0, it's at it's equilibrium position (x=0) and is moving in the positive x-direction. At 0.20 s later it is at x = 0.25 m.
a. What is the lowest possible value for the spring constant?
b. What is the second lowest possible value for the spring constant?
a. What is the lowest possible value for the spring constant?
b. What is the second lowest possible value for the spring constant?
Answers
Answered by
Damon
if 0 at 0 and v+ then
x = .3 sin 2 pi f t
.25 = .3 sin 2 pi f(.2)
.8333 = sin 2 pi f(.2)
so
2 pi f(.2) = sin^-1 .8333 = 56.4 deg = .985 radians
so
2 pi f = 4.92 radians/second
but
w = sqrt(k/m) = 2 pi f
so
k/m = (4.92)^2
k = .4 (4.92)^2
the next one will be at twice the frequency so it makes a second trip in that time
2 pi f = w = 9.84 radians /second
x = .3 sin 2 pi f t
.25 = .3 sin 2 pi f(.2)
.8333 = sin 2 pi f(.2)
so
2 pi f(.2) = sin^-1 .8333 = 56.4 deg = .985 radians
so
2 pi f = 4.92 radians/second
but
w = sqrt(k/m) = 2 pi f
so
k/m = (4.92)^2
k = .4 (4.92)^2
the next one will be at twice the frequency so it makes a second trip in that time
2 pi f = w = 9.84 radians /second
Answered by
Dan
Thanks, I wasn't able to get the second part of the answer (got a spring constant of 24.21 N/m) correct though.
Answered by
Dan
For the second part, since this is a sine function, there's a particular angle that would have the same value as 0.8333. Like sin(0 deg) and sin(180 deg) equals to 0. So I did 180 - 56.4 to get that particular angle. Once I got that, I converted it to radians and repeated the steps you provided after that. Thanks for the help Damon.