course Phy 121 {҇[苲[Student Name:
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17:35:45 `q001. Note that this assignment contains 8 questions. An early experiment in this course demonstrated that the net force restoring a pendulum to its equilibrium position was directly proportional to its displacement from equilibrium. This was expressed in the form F = - k * x, where x stands for the displacement from equilibrium and k is a constant number called the Restoring Force Constant, or sometimes a bit more carelessly just the Force Constant. A current experiment demonstrates that the motion of a pendulum can be synchronized with the horizontal component of a point moving around a circle. If the pendulum mass is m and the force constant is k, it follows that the angular velocity of the point moving around the circle is `omega = `sqrt( k / m ). If a pendulum has force constant k = 36 Newtons / meter and mass 4 kg, what is `omega? How long does it therefore take the pendulum to complete a cycle of its motion? **** we need a simulation here ****
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RESPONSE --> Force restoring a pendulum to its equilibrium position: F = -k * x Angular velocity of the point moving around the circle: `omega = `sqrt(k/m) k = 36N m = 4kg `omega = `sqrt(36N/4kg) = 3 rad/s `dt = (2 pi*rad)/(3 rad/s) = 2.09 second
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17:37:13 Since `omega = `sqrt( k / m), we have `omega = `sqrt( (36 N/m) / (4 kg) ) = `sqrt( 9 (N/m) / kg ) = `sqrt( 9 [ (kg m/s^2) / m ] / kg ) = `sqrt(9 s^-2) = 3 rad/s. Always remember that this quantity stands for the angular velocity of the point on the reference circle. [ There is a good reason why we get the radian unit here, but to understand that reason requires a very good understanding of calculus so we're not going to discuss it at this point.] A cycle of pendulum motion corresponds to a complete trip around the circumference of the circle, an angular displacement of ` pi radians. So if the reference point is moving around the circle at 3 rad/s, to complete a cycle of 2 `pi rad requires time T = 2 `pi rad / (3 rad/s) = 2 `pi / 3 sec, or approximately 2.09 sec. This time is called the Period of Motion of the pendulum, and is customarily designated T.
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RESPONSE --> I understand this problem.
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19:07:09 `q002. Recall that a pendulum with mass m and length L experiences a restoring force F = - m g / L * x, so that we have F = - k x with k = m g / L. What is the period of motion of a pendulum of length 3 meters and mass 10 kg? What would be the period of a pendulum of length 3 meters and mass 4 kg? Does your result suggest a conjecture?
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RESPONSE --> F = -m*g / L*x = -k*x k = m*g / L L = 3m m = 10kg k = (10kg * 9.8m/s^2) / 3m = 32.67 kg*m/s^2 / m = 32.67 N/m `omega = `sqrt(k/m) = `sqrt(32.67N/m / 10kg) = 1.807 rad/s T = 2 pi*rad/`omega = 2 pi*rad/1.807rad/s = 3.477 seconds L = 3m m = 4kg Since the pendulum is the same length as the previous one, it will take it the same amount of time because it will travel at the same angular velocity. The period of motion will also be 3.477 seconds.
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19:10:35 For a pendulum 3 meters long with mass 10 kg, we have k = m g / L = 10 kg * 9.8 m/s^2 / (3 meters) = 32.7 ( kg m/s^2 ) / m = 32.7 N / m. The angular velocity of the reference point for this pendulum is thus `omega = `sqrt( k / m ) = `sqrt ( 32.7 N/m / (10 kg) ) = `sqrt( 3.27 s^-2) = 1.81 rad/s. For a pendulum 3 meters long with mass 4 kg we have k = m g / L = 4 kg * 9.8 m/s^2 / (3 meters) = 13.1 N / m, so `omega = `sqrt( 13.1 N/m / (4 kg) ) = `sqrt( 3.28 s^-2) = 1.81 rad/s. These angular frequencies appear to be the same; the only difference can be attributed to roundoff errors. This common angular frequency implies a period T = 2 `pi / `omega = 2 `pi / ( 1.81 rad/s ) = 3.4 sec, approx.. Noting that both pendulums have length 3 meters we therefore conjecture that any pendulum of length 3 meters will have an angular frequency of 1.81 radians/second and period approximately 3.4 sec. We might even conjecture that the period of a pendulum depends only on its length and not on its mass.
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RESPONSE --> I understand this problem.
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19:39:35 `q003. What is a symbolic expression for the period of a pendulum of length L and mass m? Hint: Follow the same reasoning steps as in the preceding example, but instead of numbers use symbols at each step.
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RESPONSE --> With a known length, L, and mass, m, we can find the restoring force constant, k = m*g/L. After finding the constant, we can then find the angular velocity, `omega = `sqrt(k/m). `omega = `sqrt(m*g/L / m) = `sqrt(g/L) Then we can find the period of motion by the equation, T = 2 pi*rad / `omega. T = 2 pi / `sqrt (g/L) This final equation shows that the period of motion is not dependent upon mass but only on the length of the pendulum.
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19:46:41 The reasoning process went like this: We found the restoring force constant k from the length and the mass, obtaining k = m g / L. Then we found the angular frequency `omega = `sqrt( k / m ) using the value we obtained for k. Our result here is therefore `omega = `sqrt( k / m ) = `sqrt( [ m g / L ] / m ) = `sqrt( g / L ). We note that the mass divides out of the expression so that the angular frequency is independent of the mass. The period is T = 2 `pi / `omega = 2 `pi / (`sqrt ( g / L ) ) = 2 `pi `sqrt( L / g ). [ If you don't see what's going on in the last step, here are the details: 2 `pi / `sqrt( g / L ) = 2 `pi / [ `sqrt(g) / `sqrt(L) ] = 2 `pi * `sqrt(L) / `sqrt(g) = 2 `pi `sqrt( L / g ) ]. Our expression for the period is also independent of the mass. This would confirm our conjecture that the period of a pendulum depends only on the length of the pendulum and is independent of its mass.
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RESPONSE --> I understand.
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19:48:09 `q004. The frequency of a pendulum is the number of cycles completed per unit of time. The usual unit of time is the second, so the frequency would be the number of cycles per second. What is the frequency of a pendulum of length 20 cm?
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RESPONSE --> L = 20 cm = 0.2m From the previous problem we found the equation, T = 2 pi*rad / `sqrt (g/L) T = 2 pi*rad / `sqrt(9.8m/s^2/0.2m) = 2 pi*rad / 7s^2 = 0.897 seconds If there is one cycle every 0.897 seconds, then we can find how many cycles per second: cycles/second = 1 cycle/0.897s = 1.114 cycles per second.
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19:49:00 We know that the period of a pendulum is T = 2 `pi `sqrt( L / g ). Using L = 20 cm we must use g = 980 cm/s^2 in order to have compatible units in our calculation, we obtain T = 2 `pi `sqrt( L / g ) = 2 `pi `sqrt( 20 cm / (980 m/s^2) ) = 2 `pi `sqrt( .02 s^-2) = 2 `pi * .14 rad/sec = .88 sec (approx). The period represents the number of seconds required for the pendulum to complete a cycle. To obtain the frequency, which is the number of cycles per second, we take the reciprocal of the period: f = 1 / T = 1 / (.88 sec / cycle) = 1.14 cycles / sec. This pendulum will go through 1.14 complete cycles in a second.
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RESPONSE --> Ok.
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19:54:18 `q005. Early in the course the period of a pendulum was said to be related to its length by the equation T = .20 `sqrt(L), where T is in seconds when L is in cm. If we rearrange the equation T = 2 `pi `sqrt( L / g ) to the form T = [ 2 `pi / `sqrt(g) ] * `sqrt(L) and express g as 980 cm/s^2, we can simplify the factor in brackets. Do so and explain how your result confirms the equation given earlier in the course.
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RESPONSE --> When we simplify the factor in brackets, we get: 2 pi / sqrt(g) = 2 pi/sqrt(980cm/s^2) = 0.2007 s/`sqrt L (Since the denominator is in centimeters, I didn't really know how to express it so I designated it as the length in centimeters.) T = (0.2007s / `sqrt L)(`sqrt L) T = 0.2007 seconds This shows that the period of motion is expressed in seconds.
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19:55:20 The factor in brackets is [ 2 `pi / `sqrt(g) ], which becomes 2 `pi / `sqrt(980 cm/s^2) = 2 `pi / ( 31.3 `sqrt(cm) / s ) = .20 s / `sqrt(cm). The equation is therefore T = .20 s / `sqrt(cm) * `sqrt(L). If L is given in cm then `sqrt(L) will be in `sqrt(cm) and the units of the calculation will be seconds.
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RESPONSE --> I understand this problem.
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19:57:54 'q006. If we wished to construct a pendulum with a period of exactly one second, how long would it have to be?
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RESPONSE --> I am assuming that we use the same formula for period of motion from the previous problem: T = 2 pi / `sqrt (g/L) 1s = 6.283 / `sqrt(9.8m/s^2 / L) 1s^2 = 39.478 / (9.8m/s^2 / L) 9.8m / L = 39.478 L = 0.248 meters
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19:58:00 Starting with T = 2 `pi `sqrt( L / g ), we can square both sides of the equation to obtain T^2 = 4 `pi^2 * L / g. We can then multiply both sides by g / 4 `pi^2 to get L = T^2 * g / ( 4 `pi^2). Substituting 1 sec for T and 9.8 m/s^2 for g, we find that the length must be L = (1 sec)^2 * 9.8 m/s^2 / ( 4 `pi^2) = .26 m, or 26 cm.
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RESPONSE --> I understand this problem.
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19:59:11 `q007. We noted earlier that simple harmonic motion results when we have a constant mass and a restoring force of the form F = - k x. We have seen that this condition is well approximated by a pendulum, as long as its amplitude of oscillation is a good bit smaller than its length (the amplitude is the maximum distance of the pendulum from its equilibrium position). This condition is also well approximated by a mass hanging from a spring, as long as the spring is light relative to the mass and isn't stretched beyond its elastic limit (the elastic limit of a typical spring is reached when the spring is stretched so far that it won't return to its original shape after being released). If a certain light spring has restoring force constant k = 3000 N / m, and if a mass of 10 kg is suspended from the spring, what will be its frequency of oscillation?
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RESPONSE --> restoring force, F = -k*x k = 3000 N/m m = 10kg `omega = `sqrt(k/m) = `sqrt(3000N/m / 10kg) = 17.32 rad/s T = 2 pi*rad/`omega = 2 pi*rad / 17.32 rad/s = 0.363 seconds Frequency = 1/T = 1 / 0.363s = 2.757 cycles per second
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20:05:45 The angular frequency of the system is `omega = `sqrt(k / m) = `sqrt ( 3000 N/m / (10 kg) ) = `sqrt( 300 s^-2) = 17.4 rad/sec. This gives a period of T = 2 `pi rad / (17.4 rad/sec) = .36 sec, and a frequency of f = 1 / T = 1 / (.36 sec/cycle) = 2.8 cycles / sec.
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RESPONSE --> I understand this problem.
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20:11:55 `q008. In the process of designing a piece of exercise equipment, the designer needs to determine the force constant of a certain fairly strong spring. Instead of stretching the spring with a known force and measuring how much it stretches, she simply suspends the spring from the ceiling by a strong rope, ties a shorter piece of rope into a loop around the lower end of the spring, inserts her foot in the loop, puts all of her weight on that foot and bounces up and down for a minute, during which she counts 45 complete oscillations of her mass. If her mass is 55 kg, what is the force constant of the spring? Hint: first find the period of oscillation, then the angular frequency.
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RESPONSE --> f = 45cycles/ 60sec = 0.75 cycle/sec T = 1/f = 1.33 seconds m = 55kg Find restoring force constant, k. `omega = `sqrt(k/m) k = `omega^2 * m `omega = 2 pi*rad/T = 2 pi*rad/1.33s = 4.714 rad/s k = (4.714rad/s)^2 * 55kg = 1221 N/m
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20:17:04 45 cycles in 60 seconds implies a period of 60 sec / (45 cycles) = 1.33 sec / cycles. A period corresponds to 2 `pi radians on the reference circle, so that the angular frequency must be 2 `pi rad / (1.33 sec) = 4.7 rad/s, approx.. Since `omega = `sqrt( k / m ), `omega^2 = k / m and k = m * `omega^2 = 55 kg * ( 4.7 rad/s ) ^ 2 = 1200 N / m, approx.. STUDENT COMMENT: I understand how the answer was obtained and I was headed in the right direction. Another problem I had was in not knowing how the mass of the woman fit in but I think I was thinking of a pendulum where we dealt with the mass of the pendulum itself and was thinking we would need to know the mass of the spring and not the mass that was on it. INSTRUCTOR RESPONSE: *&*& In these problems we are considering ideal springs, which have negligible mass and perfectly linear force characteristics. In precise experiments with actual springs the mass of the spring does have to be considered, but this is a complex calculus-based phenomenon (for example any part of the spring experiences only the force constant of the part between it and the fixed end of the spring). *&*&
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RESPONSE --> I understand this problem.
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Ȣy{ªϡy assignment #034 034. `query 34 Physics I 07-25-2006
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00:08:50 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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RESPONSE --> For small x, the angular displacement of the pendulum equals L * angle. Since the restoring force equals -k*x, we can change a few things: x = L * angle angle = x/L F = m * g * angle = m * g * x/L Then from this equation, F = m*g*x/L, we can see that the restoring force constant is equal to m*g/L.
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00:10:00 ** 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 understand this concept.
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00:12:38 What does simple harmonic motion have to do with a linear restoring force of the form F = - k x?
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RESPONSE --> We can view this motion as a circle which includes a right triangle. I don't really know how to describe the relationship between SHM and linear restoring force without using equations. Linear restoring force, F = -km SHM, T = -(mg/L)x As you can see, these two equations with linear restoring force and SHM are related by the k value and m*g/L. Anytime we have F = -kx, we have m*d^2*x/`dt^2 which is simplified to x(t) = A cos(`sqrt(k/m)t).
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00:14:11 ** 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 --> I got totally lost on this one. I don't understand the explanation. I was trying to use Class Notes #33 for help, but I don't know if I'm even on the right track.
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00:17:46 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 --> For simple harmonic motion, the angular velocity of a point on the reference circle is represented by the equation: `omega = `sqrt(k/m) = `sqrt(-F/x / m).
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00:18:00 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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RESPONSE --> I understand this problem.
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00:22:12 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 --> `theta = `omega * t If we draw this illustration as a triangle we will see that the x coordinate is the leg of a right triangle. We will use A to represent the hypotenuse which is also the radius of the circle: x = A*cos(theta) = A * cos(`omega*t)
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00:24:29 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 --> I understand this problem.
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00:29:05 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 are trying to find the work done (the product of the force required and the displacement) which is equal to the kinetic energy required. `dW = F*`ds The force exerted on the system is an average. Since the initial force is zero, we know that the final force is -kx or -kA because x = A. Therefore the average force is -kA/2. The `dw is equal and opposite to the force done on the system, therefore, the `dW = kA/2 * `ds. Since the displacement is A, `dW = k(A^2)/2 We can read this equation as 1/2kA^2, which is very similar to 1/2mv^2. Therefore, to find the velocity, we set these two equal to each other and solve for v. 1/2 mv^2 = 1/2 kA^2 v = `sqrt(k/m) * A
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00:30:15 ** 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 --> OK. NOTE: I did not mention the potential energy in my solution, but I understand the relationship between it and the work required.
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00:32:26 Query Add comments on any surprises or insights you experienced
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RESPONSE -->
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00:32:31 as a result of this assignment.
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RESPONSE --> I had trouble explaining exactly how linear restoring force and simple harmonic motion are related.
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