course phy201 ۔|U~qRΰաGassignment #034
......!!!!!!!!...................................
12:29:53 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 --> As you said in the class notes: If the angle of the pendulum with the vertical is small, practically all of its displacement back to the equilibrium position will be in the x direction. We can therefore assume that the displacement in the y direction is small enough to be neglected.
.................................................
......!!!!!!!!...................................
12:30:15 ** 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 --> It was helpful to this this concept illustrated mathematically as well.
.................................................
......!!!!!!!!...................................
12:49:20 What does simple harmonic motion have to do with a linear restoring force of the form F = - k x?
......!!!!!!!!...................................
RESPONSE --> This equation describes how the restoring force seen with SHM is always opposite in direction to the displacement. The force and displacement change simultaneously as each cycle repeats.
.................................................
......!!!!!!!!...................................
12:57:53 ** 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 --> The calculus is N/A for me, but I do see now how I could have also included an explanation of how SHM models a reference circle, which you mathematically illustrated with A sin(`omega t) + B cos(`omega t) = C sin(`omega t + `phi)
.................................................
......!!!!!!!!...................................
12:58:50 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)
.................................................
......!!!!!!!!...................................
12:58:54 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 --> ok
.................................................
......!!!!!!!!...................................
13:03:17 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 --> As you said in the synopsis, The detailed motion can be described by the equation x(t) = A cos( omega * t ), where A is the maximum displacement from equilibrium.
.................................................
......!!!!!!!!...................................
13:03:21 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
.................................................
......!!!!!!!!...................................
13:13:36 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 --> Solve .5mv^2=.5kA^2 for v=A * `sqrt(k/m)
.................................................
......!!!!!!!!...................................
13:13:42 ** 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
.................................................
......!!!!!!!!...................................
13:14:35 Query Add comments on any surprises or insights you experienced
......!!!!!!!!...................................
RESPONSE --> Insights include understanding how SHM can be illustrated with a circle.
.................................................
......!!!!!!!!...................................
13:14:39 as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> .
.................................................
"