course Phy 121 ?????????????\?T??assignment #034????????????
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16:36:53 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?
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16:38:11 ** 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. **
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RESPONSE --> I read class notes 33 and they were really confusing. I still don't understand this explanation either.
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16:41:14 What does simple harmonic motion have to do with a linear restoring force of the form F = - k x?
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RESPONSE --> The notes refer to the rubberband example and says its pulled toward an equilibrium position. It also referred to the potential energy of the system which is .5kx^2. I have not idea how this applies to harmonic motion.
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16:42:20 ** 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. **
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RESPONSE --> ok?
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16:43:03 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?
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RESPONSE --> The angular velocity is omega = sqrt(k/m).
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16:43:08 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.
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16:44:05 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?
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RESPONSE --> The x cooridinate is x = A cos (omega*t)
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16:44:13 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)
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RESPONSE --> ok.
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16:45:27 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?
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RESPONSE --> We would have to multiply it by the equilibrium position to get the net force.
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16:46:32 ** 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.
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RESPONSE --> Oh, I see that teh KE at equilibrium is .5kA^2 which in turn is the PE. Then you can easily solve for v.
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16:47:06 Query Add comments on any surprises or insights you experienced
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RESPONSE --> I think this reference circle and harmonic motion is pretty difficult. I am trying to learn using the intro problem sets, but it is still confusing.
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16:47:10 as a result of this assignment.
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