Recent content by chriswev

I've just spoke to another instructor at the uni and he gave me the formula X = A*sin(wt+phi) I solved for phi and got 40.58 rad/s Does this look right anyone? Now I assume i put phi into the formula with A and w and leave X and t as letters?

so: 0.015 * 0.01 = 0.022cos(18.08t + phi) + 0.03 0.00015 / 0.022 = cos(18.08 * 0.01 + phi) + 0.03 0.0068 - 0.03 = cos(0.1808 + phi) -0.0232 = cos0.1808 + cos phi -0.0232 / cos1 = cos phi -0.0232 = cos phi cos -1(-0.0232) = phi phi = 91.33 rad/s ? I'm sure I've got this...

So the position i am at now is:- (I hope my earlier calculations arn't wrong) x(t) = 0.022 cos(18.08 * 0.01 + (pi / 2)) + 0.015 ? so my answer is: x(t) = 0.037 ( does this even sound reasonable?) My course notes and the books I have are of no help at all on this question. I don't...

i've now got x(t) = 0.0632 does this seem right? and is it the final answer? I'm slightly worried as i have not used the x=1.5cm?

Yeah that makes sense and was kinda the area I was thinking. I've just found this: x(t) = xm(the amplitude of motion) cos 2*pi*t/T and think this might be the one to use, not sure how this will work out though

1.5kg mass on spring streatched 3cm from natural length. oscillationg amplitude of 2.2cm. when t=0.01s, displacement x = 1.5 cm, I'm assuming gravity of 9.81 m/s I've worked out k to be 490.5 N, w to be 18.08, T to be 0.348 secs and f to be 2.97Hz, which i think are all correct. the...