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22:08:48 `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 ****......!!!!!!!!...................................
RESPONSE --> `omega = `sqrt( k / m ) so the sqrt (36N/m / 4kg) = 3m/s
I know one cycle is 2pi, and I have 'dt = 'ds/v in my notes but I don't know the radius or length of the pendulum. Maybe 2pi / 3m/s = 2.09 but I'm not sure that this is right... the units don't match up it seems..................................................
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22:10:35 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.......!!!!!!!!...................................
RESPONSE --> Ok so I did everything right but I was worried about the units because I am not gifted at calculus and can't figure out the units very easily. I see now that if I'd used rad/s for the angular velocity like I should have, then I wouldn't have wondered about getting the time.
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22:17:16 `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?......!!!!!!!!...................................
RESPONSE --> The period of motion for L = 3m and m = 4kg is: F = - mg/L*x = - 4kg*9.8m/s^2 / 3m *x = -13.07 * x. I'm not sure what the x is.
Now I'm confused... it's asking for the period of motion, and just the period. For some reason this isn't ringing a bell (though, I certainly wouldn't be surprised if I've done it a dozen times before and just don't recognize it or haven't heard it referred to this way...) As for a conjecture, I don't have any clue..................................................
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22:21:45 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.......!!!!!!!!...................................
RESPONSE --> I thought that both questions had a length of 3m and a mass of 4kg, I somehow misread the first question, which is why I think I might have been so confused. I thought there had to be different equations since I had been given the same mass and length. Turns out I just read it wrong. I think I had the wrong equation anyways, I used - mg/L*x...
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22:23:58 `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 --> Let's see... k = m g / L, which leads to 'omega = sqrt(k / m). Then for the period T = 2'pi / omega...
I think that's what this question is asking for....................................................
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22:26:22 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.......!!!!!!!!...................................
RESPONSE --> I certainly didn't go into that amount of detail, but I wasn't too far off. I find it hard to believe that a pendulum of mass 4 kg will take the same time to rotate as a pendulum with a mass many times larger (or smaller) than that.
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22:39:51 `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 --> 2 `pi `sqrt( L / g ) so 2pi sqrt(.2m / 9.8m/s^2) = 0.128m/s
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22:42:30 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.......!!!!!!!!...................................
RESPONSE --> I (as usual) forgot to square root (L/g) in my calculator though I did show it in my answer, which is why mine differs. As for f = 1/T, I would have never thought to do that.
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22:48:11 `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 = [ 2 `pi / `sqrt(g) ] * `sqrt(L) T = (2pi / sqrt980cm/s^2) * sqrt(20cm) T = 0.2 cm/s^2 * 4.47cm = 0.89s
So again for the frequency, 1 / .89s = 1.14cycles/sec, which is the same as T = .2'sqrt(L)....................................................
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22:48:36 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.......!!!!!!!!...................................
RESPONSE --> ok
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22:50:08 '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 --> T = .2sqrt(L) L = (1cycle/sec*.2)^2 length = .04cm
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22:51:09 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.......!!!!!!!!...................................
RESPONSE --> ok well I used the approximate value equation.
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`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?......!!!!!!!!...................................
RESPONSE --> K = 3000N/m and m = 10kg.
I'm not sure if I'm using the right methods here but this is my attempt. Omega = sqrt(K/m) = sqrt (3000N/m / 10kg) = 17.32 rad/s T = 2pi / omega = 2pi / 17.32rad/s = 0.36sec The frequency is f = 1 / T. So 1 / 0.36s = 2.78cycles/second..................................................
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10:53:21 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.......!!!!!!!!...................................
RESPONSE --> ok cool.
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11:06:52 `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 --> We know that m = 55kg, T = 60sec, and that there's 45 oscillations.
I've tried to find the period of oscillation, but I need to know the angular frequency first... But the hint said to find that last so I'm not really sure how to do this. The period of oscillation is T = 2pi rad / omega. 60s = 2pi rad / omega. omega = 2pi rad / 60s = .105 rad/s angular frequency: omega = sqrt (k/m) = sqrt(mg/L / m) Yeah for some reason I'm pretty lost here. I don't think it's difficult to do, but I just can't seem to figure it out at the moment..................................................
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11:09:04 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). *&*&......!!!!!!!!...................................
RESPONSE --> Ok so I didn't need to focus on the mass like I was trying to. I was thinking of what I could use the 45 oscillations for, but I just didn't consider them to be cycles, though I'm not sure how I didn't realize that.
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