Phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
I wanted to clear this point up before the final exam:
I know that the force of a pendulum is m(g)*(displacement/length). Once I have this I am confused as how to find velocity. Would I find work which would be F*distance with distance being .1362m then use the equation for KE to find velocity? Or would I calculate angles using trigonometry and use an equation?
** **
** **
The average force between the origin and the position of the pendulum would be half the force you give above, i.e., 1/2 m g x / L (this is obtained by averaging the force at the origin, which is 0, with the force at position x).
The potential energy of the pendulum, relative to equilibrium, is therefore F_ave * `ds = (1/2 m g x / L) * x= 1/2 (m g / L) * x^2.
In general if restoring force constant is k then the energy at position x is F_ave * `ds = (1/2 k x) * x = 1/2 k x^2.
If the amplitude of motion is A, then the maximum PE of the oscillation is 1/2 k A^2. The PE change from position A to position x is 1/2 k A^2 - 1/2 k x^2, and this is the KE at position x.
Phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
I have having trouble with this question so i was wondering if you could check my solution I cant find any PS questions that address this:
Problem Number 2
A simple harmonic oscillator of mass 40 kg has a period of .04 seconds.
* If the amplitude of its motion is 16560 meters, what are the maximum magnitudes of its acceleration and velocity?
* At what displacements from equilibrium can each maximum occur?
OK so first i found omega to be (2pi)/.04sec which gave 157rad/sec.
Once i have omega I used the equation v(t)=-omegaA(sin(omega*t)
I am pretty sure that the max velocity is omega(A) which would be 2599920m/s. Then I found acceleration by finding dt. velocity changed from 0 to 2599920m/s in t=.03sec. I found this by finding when sin(omega*t)=-1. So dividng dv/dt of .03 sec would give acceleration of 86664000m/s^2.
I am guessing that the displacements from equilibrium would just be the dt of .03, but I am not sure what else i would solve for here.
The reference-circle picture consists of a point moving around a reference circle of radius 16560 meters with angular velocity 50 pi rad / sec (approximately equal to your 157 rad/s). The velocity of the reference point is
v = 16560 m * 50 pi rad/s = 2 600 000 m/s, approx, or about 2.6 * 10^6 m/s.
and its centripetal acceleration is
a_cent = v^2 / r = (2.6 * 10^6 m/s)^2 / (16560 m) = 4.1 * 10^8 m/s^2, approx..
Alternativel this result is
a_cent = r omega^2 = 16560 m * (156 rad/s)^2 = 4.1 * 10^8 m/s^2, approx.
The oscillator is modeled by projecting the reference point onto any straight line thru the origin (e.g., through the x or y axis, but any other point would also give a model).
The maximum velocity of the oscillator occurs when its direction of motion matches that of the reference point, and is at these points equal to the velocity of that point. The maximum velocity is therefore 2.6 * 10^6 m/s.
The maximum acceleration of the oscillator occurs when the direction of the centripetal acceleration vector is parallel to the direction of motion, and is at these points equal to the centripetal acceleration of the reference point. The maximum acceleration of the oscillator therefore has magnitude 4.1 * 10^8 m/s^2.
** **
** **
I am hoping you can let me know if this work is correct or direct me to any PS that would help me with this. Also if there are errors could you point them out.