#$&* course PHY 201 9:30 PM on 5/7/13 034. Simple Harmonic Motion*********************************************
.............................................
Given Solution: Since `omega = `sqrt( k / m), we have `omega = `sqrt( (36 N/m) / (4 kg) ) = `sqrt( 9 (N/m) / kg ) = `sqrt( 9 [ (kg m/s^2) / m ] / kg ) = `sqrt(9 s^-2) = 3 rad/s. Always remember that this quantity stands for the angular velocity of the point on the reference circle. [ There is a good reason why we get the radian unit here, but to understand that reason requires a very good understanding of calculus so we're not going to discuss it at this point.] A cycle of pendulum motion corresponds to a complete trip around the circumference of the circle, an angular displacement of ` pi radians. So if the reference point is moving around the circle at 3 rad/s, to complete a cycle of 2 `pi rad requires time T = 2 `pi rad / (3 rad/s) = 2 `pi / 3 sec, or approximately 2.09 sec. This time is called the Period of Motion of the pendulum, and is customarily designated T. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Recall that a pendulum with mass m and length L experiences a restoring force F = - m g / L * x, so that we have F = - k x with k = m g / L. What is the period of motion of a pendulum of length 3 meters and mass 10 kg? What would be the period of a pendulum of length 3 meters and mass 4 kg? Does your result suggest a conjecture? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: k = mg / L = 10kg(9.8 m/s^2) / 3 m = 32.7 N / m 'omega = sqrt(k / m) = sqrt(32.7 N / m / 10 kg) = 1.8 rad / s T = 2 pi rad / 'omega = 2 pi / 1.8 rad/s = 3.5 s k = 4kg(9.8 m/s^2) / 3 m = 13.1 rad/s T = 2 pi / 13.1 rad/s = 0.48 confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = 2 pi / 'omega = T = 2 pi / (sqrt(k / m)) T = 2 pi (sqrt((mg/L) / m) T = 2 pi (sqrt(g / L) confidence rating #$&*:#8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. NEARLY CORRECT STUDENT SOLUTION: k = mg/Lomega = sqrt(k/m)period of pendulum = 2pi rad / omega INSTRUCTOR RESPONSE: Your expressions are correct, but your answer isn't in terms of L and m. Using your expressions, it's easy to get a result in terms of L and m. k = m g / L soomega = sqrt(k / m) = sqrt( (m g / L) / m) = sqrt( g / L). So period of pendulum = 2pi rad / omega = 2 pi / (sqrt(g / L ). This answer would be acceptable for a Physics 121 student, but the denominator involves a fraction, so it isn't quite in standard form. 2 pi / (sqrt(g / L ) = 2 pi * sqrt( L / g), which is standard form. STUDENT QUESTION: I knew you had to obtain omega somehow. I got a little confused on how omega was equal to sqrt(g/L). Could you explain that a little more INSTRUCTOR RESPONSE: The net force on a pendulum displaced x units from equilibrium, where | x | < < L (magnitude of displacement is much less than pendulum length), is F_net = - (m g / L) * x. • This was found in previous assignment, using the fact that the vertical component of pendulum tension is very close to m g, and is in the same proportion to the horizontal component as the displacement to the length of the pendulum. Thus F_net = - k x, where k = m g / L. Since omega = sqrt(k / m), we have omega = sqrt( (m g / L) / m) = sqrt(g / L). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = sqrt(g / L) = sqrt(9.8 m/s^2 / 0.2 m) = 7 s f = 1 / T = 1 / 7 s = 0.14 cycles per second or Hz. confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. STUDENT COMMENT: i didn't know where to start here, i see how you got the answer for T, but had no idea to get frequency by dividing 1 by T INSTRUCTOR RESPONSE: Be sure to make a note of this. It's not difficult to remember, but if you think in terms of the meanings of the quantities it's easier than trying to keep a bunch of similar-looking formulas straight. Recall the definitions of frequency and period from the very first experiment of the course, the Introductory Pendulum Experiment. It was not an accident that we started the course with the very last, and perhaps most challenging topic. We have encountered the pendulum at many points in the course, and these last few assignments are designed to complete our understanding of this common system. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Forgot to use the rad notation. 2 pi * 0.14 = 0.879 sec, f = 1 / T = 1 (0.879 sec / cycle) = 1.14 cycles per sec ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = [ 2 pi / 'sqrt(980 cm/s^2] * 'sqrt(L) T = [0.20 cm / s] * 'sqrt(L) This matches the equation 0.20 'sqrt(L). confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. STUDENT COMMENT: this factoring and making equations confuse me, it is hard enough to figure out the number to go into given equations INSTRUCTOR RESPONSE: Such confusion is typical among Principles of Physics students, is not unusual among General College Physics and even occurs among students in University Physics. When the expressions get complicated, be sure you write them out on paper. You can't hope to read or operate on a challenging algebra expression without writing it down in standard form. The algebra involved here approaches but does not exceed the level of a high school Algebra II class and should be accessible to students in all courses. However students rarely master their last mathematics class, and Principles of Physics students whose mathematical background goes only through Algebra II will typically have some difficulty with the more challenging algebra, such as that in these problems. Though nothing this algebraically challenging occurs on Principle of Physics tests, it is useful to try to understand the algebraic thinking used here. General College Physics students are expected to do algebra at this level, and University Physics students are expected to have full mastery of basic algebra as well as a workable knowledge of basic calculus. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. If we wished to construct a pendulum with a period of exactly one second, how long would it have to be? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = sqrt(g / L) 1 s = sqrt(980 cm/s^2 / L) (1 s)^2 = ((980 cm/s^2) / L)^2 L = 31.3 confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. STUDENT QUESTION: why did you sq both sides in the begining, i had the right idea and the right equation except for the 4pi rad INSTRUCTOR RESPONSE: L occurs inside a square root. You can't solve for L if L is inside the square root. You wish to solve for L. If you square sqrt(L/g) you get (sqrt(L/g))^2 = L/g. If you square just one side of an equality, then that side changes and the other doesn't, so you no longer have an equality. If you square both sides of an equation, a solution to the original equation is also a solution of the resulting equation. So we square both sides of the equation. Having done so, it becomes fairly easy to solve for L. STUDENT QUESTION I don't know where to go from here. i don't know how to calculate the frequency where did this equation come from : f = 1 / T = 1 / (.36 sec/cycle) = 2.8?? is it related to F = - k x somehow? INSTRUCTOR RESPONSE You don't really need an equation to change between frequency and period. All you need is to understand how the two are related. Once you understand this, the relationship is very simple. T is the time required for 1 cycle. f is the number of cycles per unit of time. You calculated period and frequency in the early pendulum experiment. In that case frequency was counted in cycles / minute. From that you obtained the number of seconds per cycle (the period) and the number of cycles per second (the frequency in cycles/second rather than cycles/minute). The frequency is the reciprocal of the period. For example if a pendulum requires a period of 2 seconds to complete a cycle, then its frequency is 1/2 cycle / second (it completes half a cycle in a second). If it requires only, say, .4 seconds to complete a cycle, then it will complete 1/.4 = 2.5 cycles every second, so its frequency is 2.5 cycles/second. Once we understand the meaning of frequency and period, we easily understand why f = 1 / T (and of course why T = 1 / f). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): If I added 2 pi to the equation, then I would have gotten the same answer. ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: omega = sqrt(k / m) = sqrt(3000 N / m / 10 kg) = 17.3 rad / s T = 2 pi rad / 'omega = 2 pi rad / 17.3 rad/s = 0.36 s f = 1 / T = 1 / 0.36 s = 2.75 cycles per second. confidence rating #$&*:32; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. STUDENT COMMENT: i divided 45by60 and not the other way, which was mistake 1, i didn't get an angular frequency and didn't know to use k = m* omega^2, i used k=mg/L INSTRUCTOR RESPONSE k = m g / L is the restoring force constant for a simple pendulum of mass m and length L. k by itself is not related to angular frequency or frequency of oscillation. The key concept here is the angular frequency omega, which is the angular velocity of the point on the reference circle. In the present problem k = m g / L is not relevant at all, since the situation does not involve a pendulum. The relationship required to analyze frequencies is • omega = sqrt(k / m). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The period is seconds per cycle and the reciprocal is frequency which is cycles per second. The test is done for one minute, which is equal to 60 seconds. 60 seconds / 45 cycles = 1.33 seconds per cycle T = 2 pi / 'omega => T = 2 pi / (sqrt(k / m)) (sqrt(k/m)^2 = 2^2 pi^2 / T^2 k = (39.5 / 1.8 s/c) * 55 kg = 1207 N / m confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 45 cycles in 60 seconds implies a period of 60 sec / (45 cycles) = 1.33 sec / cycle. 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). *&*& STUDENT COMMENT I was obviously confused about how to go from 45 oscillations in 60 seconds, and make that into the period T. I was also confused as to why your period was in seconds / cycle. That just seems strange to me. INSTRUCTOR RESPONSE The period of an oscillation is the time required for a cycle. If time intervals are measured in seconds, then, the period would be the number of seconds per cycle. The unit would be seconds / cycle, though the 'cycle' part is pretty much understood and could be left off. Leaving it off we have the unit 'seconds'. Similarly the frequency of an oscillation is expressed in cycles/second. The 'cycles' is understood and could be left off, so that the unit is 1/second or (second)^(-1). Basically, then, the word 'cycle' isn't officially part of any unit, but can be used as appropriate to clarify the meaning of either period or frequency. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q009. When loaded with a mass of 5 kg, the length of a certain ideal spring increases by 8 cm. The system is then pulled down an additional 10 cm and released, with the result that the mass undergoes simple harmonic motion. What is the force constant of the spring? What therefore is the angular frequency of the motion? What is the frequency of the oscillation of the mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I am confused by the second length of 10 cm. Do we solve for the 5 kg and 8 cm first, but after that, i am not sure what to do with the 10 cm. k = mg / L = 5kg(9.8 m/s^2) / 0.8 m = 61.3 N / m 'omega = sqrt(k / m) = sqrt(61.3 N / m / 5 kg) = 3.5 rad / s T = 2 pi rad / 'omega = 2 pi / 3.5 rad/s = 1.8 seconds per cycle f = 1 / T = 1 / 1.8 s = 0.56 cycles per second confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!