Did you get my emails about the SHM program not working and about me taking test 2? I haven't heard anything back so I wanted to make sure you received them." "Ոֿ{sYHz assignment #034 ~ȪsJ콥 Physics I 12-20-2005
......!!!!!!!!...................................
00:28:55 Query Class Notes #33 Why do we say that a pendulum obeys a linear restoring force law F = - k x for x small compared to pendulum length?
......!!!!!!!!...................................
RESPONSE --> For small displacements the pendulum obeys the F = - k x law. F is the restoring force Tx, and k is the quantity m g / L.
.................................................
......!!!!!!!!...................................
00:33:03 ** The vertical component of the tension in the string is equal to the weight m * g of the pendulum. At angle `theta from equilibrium we have T cos(`theta) = m * g so T = m * g / cos(`theta).
The horizontal component of the tension is the restoring force. This component is T sin(`theta) = m * g * sin(`theta) / cos(`theta) = m * g * tan(`theta). For small angles tan(`theta) is very close to `theta, assuming `theta to be in radians. Thus the horizontal component is very close to m * g * `theta. The displacement of the pendulum is L * sin(`theta), where L is pendulum length. Since for small angles sin(`theta) is very close to `theta, we have for small displacements x = displacement = L * `theta. Thus for small displacements (which implies small angles) we have to very good approximation: displacement = x = L `theta so that `theta = x / L, and restoring force = m * g * `theta = m * g * x / L = ( m*g/L) * x. **......!!!!!!!!...................................
RESPONSE --> Yeah, I'm having trouble following this. I understand everything and the algebra behind it, but I still hardly understand what the question's even asking for and am not sure I could apply this answer to a question involving actual data.
.................................................
......!!!!!!!!...................................
00:37:32 What does simple harmonic motion have to do with a linear restoring force of the form F = - k x?
......!!!!!!!!...................................
RESPONSE --> I know the class notes compare the situation that used a rubber band, where the force was -kx and was used to pull an object back toward the equilibruim position. It also says that x was 1/2kx^2.
.................................................
......!!!!!!!!...................................
00:39:39 ** the essential relationship is F = - k x; doesn't matter if it's a pendulum, a mass on an ideal spring, or any other system where net force is a negative constant multiple of the displacement from equilibrium.
F = m * a = m * x'', so F = - k x means that m * x'' = - k x. The only functions whose second derivatives are constant negative multiples of the functions themselves are the sine and cosine functions. We conclude that x = A sin(`omega t) + B cos(`omega t), where `omega = `sqrt(k/m). For appropriate C and `phi, easily found by trigonometric identities, A sin(`omega t) + B cos(`omega t) = C sin(`omega t + `phi), showing that SHM can be modeled by a point moving around a reference circle of appropriate radius C, starting at position `phi when t = 0. **......!!!!!!!!...................................
RESPONSE --> I think it's obvious that I really have no idea at what it is I'm supposed to be doing to figure out the answers to these questions.
.................................................
......!!!!!!!!...................................
00:41:50 For a simple harmonic oscillator of mass m subject to linear net restoring force F = - kx, what is the angular velocity `omega of the point on the reference circle?
......!!!!!!!!...................................
RESPONSE --> omega = sqrt (k / m)... where k = mg/L.
.................................................
......!!!!!!!!...................................
00:44:04 STUDENT RESPONSE: omega= sqrt (k/m)
INSTRUCTOR COMMENT: Good. Remember that this is a very frequently used relationship for SHM, appearing in one way on another in a wide variety of problems.......!!!!!!!!...................................
RESPONSE --> Yay for something I might actually understand.
.................................................
......!!!!!!!!...................................
00:44:56 If the angular position of the point on the reference circle is given at clock time t by `theta = `omega * t, then what is the x coordinate of that point at clock time t?
......!!!!!!!!...................................
RESPONSE --> x = A cos(omega*t)
I think that's right....................................................
......!!!!!!!!...................................
00:45:19 since theta=omega t, if we know t we find that x = radius * cos(theta) or more specifically in terms of t
x = radius*cos(omega*t)......!!!!!!!!...................................
RESPONSE --> ok.
.................................................
......!!!!!!!!...................................
00:57:32 Query introductory problem set 9, #'s 1-11 If we know the restoring force constant, how do we find the work required to displace the oscillator from its equilibrium position to distance x = A from that position? How could we use this work to determine the velocity of the object at its equilibrium position, provided we know its mass?
......!!!!!!!!...................................
RESPONSE --> I think we start off finding the angular velocity, sqrt(k/m), then we multiply omega by the time period and find the y position. After that, we find the average velocity, 'dy/dt and use KE = .5 m v^2 to find the work required.
To determine the velocity at equilibrium, I think using v = sqrt(KE/m) would work but I'm not sure if this applies here or not. This is based off a poor understanding of Problem set 9, # 6... I couldn't figure out the very first calculation so I tried to use it to answer this question but am not sure it's right..................................................
......!!!!!!!!...................................
01:00:19 ** You can use the work 1/2 k A^2 and the fact that the force is conservative to conclude that the max PE of the system is 1/2 k A^2. This PE will have transformed completely into KE at the equilibrium point so that at the equilibrium point KE = .5 m v^2 = .5 k A^2. We easily solve for v, obtaining
v = `sqrt(k/m) * A. **{}{}STUDENT COMMENT: I'm a little confused by that 1/2 k A^2.{}{}INSTRUCTOR RESPONSE: {}{}That is the PE at x = A. To directly reason out the expression PE = .5 k A^2 we proceed as follows:{}{}PE = work done by system in moving from equilibirum * displacement = fAve * `ds.{}{}The force exerted on the system at position x = A is -k A. The force exerted at position x = 0 is 0. Force changes linearly with position. So the average force exerted on the system is {}{}( 0 - kA) / 2 = -1/2 k A.{}{}The force exerted by the system is equal and opposite, so {}{}fAve = 1/2 k A.{}{}The displacement from x = 0 to x = A is `ds = A - 0 = A. {}{}We therefore have{}{}PE = fAve `ds = 1/2 k A * A = 1/2 k A^2.{}{}This is also the area beneath the F vs. x graph between x = 0 and x = A. That region is a triangle with base A and altitude k A so its area is 1/2 base * height = 1/2 k A^2.......!!!!!!!!...................................
RESPONSE --> Physicists amaze me.
.................................................
......!!!!!!!!...................................
01:05:33 Query Add comments on any surprises or insights you experienced
......!!!!!!!!...................................
RESPONSE --> I always attempt the problem set questions (sadly attempt actually...) but a lot of the time, I'm trying to follow the explanations and my calculations won't match up. For example, in Problem set 9, #6, the angular velocity is sqrt(930N/m / 7kg). I understand this formula, but I cannot for the life of me figure out where 930N/m came from.
Outside of the first few questions at the beginning of the query, I understand the problem set questions and equations for the most part. I'm still not able to get the SHM program to work so I'm at a standstill with the labs. Does experiment 32 use the same program? The instructions aren't very long and I'm not exactly sure about what it is I need to submit so I'm thinking it has something to do with that program..................................................
......!!!!!!!!...................................
01:05:38 as a result of this assignment.
......!!!!!!!!...................................
RESPONSE -->
.................................................
"