course PHY 231 This is the rest of the work for PHY 231 assignments 34 and 35. Kgԉky锫yassignment #034
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21:54:09 `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 --> We are given the formula `omega = `sqrt( k / m We are given k = 33 N/m and m = 4kg So omega = sqrt( (36 N/m) / (4kg)) omega = sqrt( (9N/m/kg)) omega = 9 rad/sec Before I will go on I will note where I am confused. Omega is measured in rad/sec. I'm not sure how we go from N/m/kg to rad/sec. So something happens when you take the square root of N/m/kg. We know the relationship omega = 2 pi / T T is the period. Since we have omega we can easily solve for T. T = 2pi / omega T = 2pi / 9rad/sec so T = 0.22 pi sec or 0.698 sec confidence assessment: 3
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21:55:47 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 do not remember if I took the square root of 9. If I didn't I want to clarify that I know it would be 3rad/sec. This in turn would make my period incorrect. self critique assessment:
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15:36:26 `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 --> The period equation is T = 2 pi * sqroot(m/k) First we need to solve for k, which is m *g/l so k = 4kg * (9.8m/sec^2 / 3 m) k = 4kg * 3.267 1/sec^2 k = 13.067 kg/sec^2 So let's solve for T since we already have m T = 2 *pi * sqroot(m/k) T = 2pi * sqroot(4kg / 3 m) T = 7.255 sec confidence assessment: 2
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15:37:29 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 messed up in the beginning when I put in the incorrect numbers for mass. self critique assessment: 2
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15:38:15 `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 --> T = 1/f = 2pi / omega = 2pi (sqroot(m/k)) confidence assessment: 2
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15:38:31 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 --> ok self critique assessment: 2
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15:42:48 `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 --> Frequency = 1/T ( T is the period) I am not sure how I can solve for this one without further information. If I knew how long it took the 20cm pendulum to complete one revolution I could just take the reciprocal of that and have my frequency. I do not know how fast it is traveling. If one period was 2 seconds, then the frequency is 1/2 seconds or 1/2 Hertz. Frequency is measured in reciprocal seconds. confidence assessment: 1
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15:46:57 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 --> I knew the formula; I was not thinking. Earlier I wrote that T = 2pi (sqroot(m/k) If we substitute the value of k we get: T = 2pi (sqroot (m / (m(g/L) With this mass cancels out and we have the square root of L/g to balance out. T = 2pi (sqroot(L/g) In this case it is easy to solve. T = 2pi (sqroot (20cm/980cm/sec^2)) T = 2 pi (sqroot(0.020408 1/sec^2) T = 2pi 0.14287 1/sec or T = 0.8976 1/sec self critique assessment: 2
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15:50:26 `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 --> T = 2pi sqroot(g) * sqroot(L) T = 2pi sqroot(980cm/sec^2) * (sqroot(20cm)) T = 2pi * 31.30 * 4.472 1/sec T = 2pi * 139.977 1/sec T = 879.51 1/sec confidence assessment: 1
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15:50:35 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 --> self critique assessment:
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15:57:40 `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 --> Now I can use the equation correctly from earlier. T = 2pi * sqroot(L/g) We have to get L by itself. First I square the entire equation to have: T^2 = 4pi^2 * L/g Now I multiply by g so that T^2*g = 4pi^2 * L Divide by 4pi^2 so that L = (T^2*g) / 4pi^2 Now I simply put in my values. The square of 1 is 1. L = (1sec^2 * 9.8m/sec^2) / (39.478m) L = 9.8m / 39.478m L = 0.248 m I can check my work by plugging in the value of L into the original equation and getting T = 1. I did and I got 1. My answer is correct. My one question is I am not sure how I am left with meters. When I divide m / m then I should be left with nothing. How does m stay? confidence assessment: 3
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15:57:59 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. Note that we would have obtained 26 cm directly if we had used g = 980 cm/s^2. The units chosen for g depend on the units we want to get for our result. STUDENT QUESTION: Why didn't we use the equation T = 0.2 'sqrt (L) for this? INSTRUCTOR RESPONSE: 0.2 is the approximate value of 2 pi / g, when L is in cm. That approximation comes from this equation. We're using the accurate equation now. The approximation was more than accurate enough for experiments, but when dealing with problems involving simple harmonic motion we don't use that approximation.
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RESPONSE --> self critique assessment:
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16:05:11 `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 --> My plan is to use the original formula for period, and then take the reciprocal of that. So T = 2pi * sqroot(m/k) T = 2pi * sqroot(10kg/3000N/m) T = 2pi * sqroot(0.00333 kg/Nm) T = 2pi * 0.0577 T = 0.3627 sec f = 1/T f = 1/0.3627 So f = 2.757 Hz I asked this earlier. I am not sure what to do with the units of this problem when I take a square root. For example, what is the square root of a meter) Or does nothing happen? I took the square root of a kg /N/m or kg / (kgm/sec^2 )*m with how it is structered I am left with 1/sec^2 confidence assessment: 2
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16:05:34 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 --> okay. I did the problem differently but it still seems to work out. self critique assessment: 2
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16:10:31 `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 --> If oscillations f = 45, we can find period T by taking the reciprocal. T = 1/f T = 0.022seconds T = 2pi / omega omega = 2pi / T omega = 2pi / 0.022sec omega = 285.599 rad/sec Now we use the following equation to find k. omega = sqroot (k/m) omega ^2 = k/m omega ^2 *m = k (285.599rad/sec)^2 * 55kg = k k = 4486183.819 N/m That seems like a very large number, but I think I have the correct steps down. confidence assessment: 1
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16:11:40 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 thought oscillations was the frequency. That started my answer in the wrong direction. self critique assessment: 2
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