utorstudent
Sep26-04, 01:59 PM
This problem set is for my oscillations and wave class; I just can't seem to make any solutions work.
The problem is a mass on spring pendulum with some given coordinates
time(s) displacement(cm) velocity(cm/s) acceleration(cm/s^2)
1.31 4.00 0 not recorded
3.67 not recorded not recorded 7.93
what length must the string be? You may need to solve this graphically
that's the question so I started by writing out the equation of motion for a simple harmonic oscillator
x(t) = Acos(wt + phi), w= omega and phi= phase shift and A is max amplitude. However the velocity is 0 at the endpoints and that happens at t=1.31 so A=4
x(t) = 4cos(wt + phi)
then I tried substituing in the coordinate restrictions
x(1.31) = 4= 4cost(1.31w +phi)
0= 1.31w + phi
phi= -1.31w
Then to get more equations I differentiated and used the restriction on acceleration
x'(1.31) = -4w sin( w1.31 + phi) =0
x'' (3.67) = 7.93 = -4w^2cos(w1.31+ phi)
which would mean
w = 1.40801
then by the simple harmonic equation
x''(t) + w^2x(t) = 0
x''(3.67) + 1.9825 x(3.67) = 0
x(3.67) = -4
so that would mean that the bob is at the other end and we could fill in the rest of the chart
time(s) displacement(cm) velocity(cm/s) acceleration(cm/s^2)
1.31 4.00 0 -7.93
3.67 -4.00 0 7.93
at this point I'm stuck. I don't really know how to finish the question but I feel it is in the right direction. Any help would be greatly appreciated.
The problem is a mass on spring pendulum with some given coordinates
time(s) displacement(cm) velocity(cm/s) acceleration(cm/s^2)
1.31 4.00 0 not recorded
3.67 not recorded not recorded 7.93
what length must the string be? You may need to solve this graphically
that's the question so I started by writing out the equation of motion for a simple harmonic oscillator
x(t) = Acos(wt + phi), w= omega and phi= phase shift and A is max amplitude. However the velocity is 0 at the endpoints and that happens at t=1.31 so A=4
x(t) = 4cos(wt + phi)
then I tried substituing in the coordinate restrictions
x(1.31) = 4= 4cost(1.31w +phi)
0= 1.31w + phi
phi= -1.31w
Then to get more equations I differentiated and used the restriction on acceleration
x'(1.31) = -4w sin( w1.31 + phi) =0
x'' (3.67) = 7.93 = -4w^2cos(w1.31+ phi)
which would mean
w = 1.40801
then by the simple harmonic equation
x''(t) + w^2x(t) = 0
x''(3.67) + 1.9825 x(3.67) = 0
x(3.67) = -4
so that would mean that the bob is at the other end and we could fill in the rest of the chart
time(s) displacement(cm) velocity(cm/s) acceleration(cm/s^2)
1.31 4.00 0 -7.93
3.67 -4.00 0 7.93
at this point I'm stuck. I don't really know how to finish the question but I feel it is in the right direction. Any help would be greatly appreciated.