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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): ok ------------------------------------------------ Self-critique rating:3 ********************************************* 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: ok confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique rating:3 ********************************************* 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: ok confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique rating:3 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: K= mg/L= 5kg*9.8m/s^2/0.18m= 272.2N/m
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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): It looks like I used the less accurate version, which made my answer off by 1 cm. ------------------------------------------------ Self-critique rating: 2 ********************************************* 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: I started by solving for the period of motion: `omega = `sqrt k/m `omega = `sqrt 3000 Nm/10 kg `omega = `sqrt 300 `omega = 17.3 rad/s Then I solved for elapsed time: `dt = 2`pi/`omega `dt = 2`pi/17.3 rad/s `dt = .363 seconds/ cycle To find the cycles per second, I took the reciprocal: 1/.363 sec/cycle = 2.75 cycles/second confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique rating: 2 ********************************************* 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: `omega = `sqrt k/m 45 oscillations/minute = .75 oscillation/s .75 oscillations/s = `sqrt k / 55kg Square both sides .5625 osc^2/s^2 = k/55kg Multiply both sides by 55 kg 30.94 = K confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): The good news was that I had some of the right ideas. The bad news was that I applied them in the wrong order, which gave me wrong answers and also showed that I had no idea what I was supposed to be doing. The given solution makes sense. ------------------------------------------------ Self-critique rating: 2 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: All I can figure out is that 5 kg of mass yields 8 cm of change. F = m * a F = 5 kg * 9.8 m/s^2 F = 49 N `dw = F * `ds `dw = 49 N * 8 cm `dw = 392 JOules I don't think any of this is relevant to what I'm supposed to be finding, but it's what I thought I saw. To find the period of motion, the formula is `omega = `sqrt k/m. Of these 3 quantities, all I know is mass = 5 kg. To find the angular frequency of motion, the formula is 2 `pi radians/`dt. I could multiply 2 `pi, but don't know `dt. To find the frequency of oscillation, I need to find cycles per second and then take the reciprocal, but I don't know where to get that information from the data given..... confidence rating #$&*:5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!