course phy201 xκD|~vĩassignment #034
......!!!!!!!!...................................
10:24:58 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 --> Because the pendulum swings in an arc, for the equation to be a good approximation, the displacement (x) must be small compared to the length (ie, small theta) or error will be too great when assuming linear displacement.
.................................................
......!!!!!!!!...................................
10:26:04 ** 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 --> ok
.................................................
......!!!!!!!!...................................
10:29:13 What does simple harmonic motion have to do with a linear restoring force of the form F = - k x?
......!!!!!!!!...................................
RESPONSE --> The object moves back and forth through it's equilibrium position trading KE for PE and back again with it's period (or cycles per unit time) determined by the spring constant and the amount of displacement but not by the mass.
.................................................
......!!!!!!!!...................................
10:30:16 ** 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 --> ok
.................................................
......!!!!!!!!...................................
11:06:57 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 = 2pi*f*cosA
.................................................
......!!!!!!!!...................................
11:08:11 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 --> I knew this but don't see how it relates omega to the position on the reference circle?
.................................................
......!!!!!!!!...................................
11:10:06 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 = Cos theta
.................................................
......!!!!!!!!...................................
11:10:23 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
.................................................
......!!!!!!!!...................................
11:16:19 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 --> dW = A * k
.................................................
......!!!!!!!!...................................
11:17:02 ** 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 --> ok
.................................................
......!!!!!!!!...................................
11:17:17 as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> ok Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
.................................................