course Phy 201 ????????a??X????assignment #034
......!!!!!!!!...................................
09:49:49 `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 K=36N/m m= 4kg 36N/m/4kg= 9 M^2/s^2 take the square root= 3m/s confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:53:12 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 --> Should have canceled the ms out to get 3 rad/s= angular velocity of the point on a reference circle. Angular displacement= 2 pi rad T= 2 pi rad/ 3 rad/s= 2pi/3s= 2.09s= Period of Motion self critique assessment: 2
.................................................
......!!!!!!!!...................................
10:07:25 `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 --> K=mg/L omega= sqrt of K/m K= 10kg*9.8m/s^2/(3m)= 326 N/m sqrt of 326N/m /10m= 5.72 rad/s T= 2 pi rad/5.72 rad/s=1.10 s K=mg/L K= 4 kg * 9.8m/s^2/ 3m K= 13.07 N/m sqrt of 13.07 N/m/ 4kg= 1.808 rad/s T= 2 pi rad/1.808 rad/s =1.11 s The pendulums of the same length took about the same time even though the masses were different confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:13:57 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 --> When multiplying 10 *9.8 should have gotten 98, must have used 980. K=mg/L= 32.7N/m omega=`sqrt(k/m)= 1.81 rad/s For 3 m long and 4 kg= k=mg/L= 4kg*9.8m/s^2/3m= 13.1N/m omega= sqrt(13.1N/m/4kg)=1.81 rad/s The angular frequencies appear to be the same. T= 2 pi/omega= 2 pi /1.81 rad/s=3.4s Becuase the pendulum lengths were the same the angular frequency and the period turned out the same even with differing masses. self critique assessment: 2
.................................................
......!!!!!!!!...................................
10:17:04 `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.
......!!!!!!!!...................................
RESPONSE --> omega= sqrt of K/M omega= sqrt(mg/L/m) The ms cancel out omega= sqrt of g/L T= 2 pi rad/ sqrt of g/L confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:19:04 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 --> T= 2 pi / omega= 2 pi/sqrt of g/L= 2 pi sqrt of L/g The period of a pendulum depends only on the length of the pendulum self critique assessment: 2
.................................................
......!!!!!!!!...................................
10:26:24 `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?
......!!!!!!!!...................................
RESPONSE --> Frequency= Number of cycles per unit of time L= .2m T= 2 pi /sqrt of g/L T= 2 pi/sqrt(.9m/s^2/.2m)= .89s Frequency = 1cycle/.89s=1.12cycles/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:27:23 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 --> F= 1/T= 1/.88 s/cycle= 1.14 cycles /s self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:34:28 `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.
......!!!!!!!!...................................
RESPONSE --> T= .2* sqrt of L T= (2 pi/ `sqrt(g)) * sqrt(L) 2 pi= 6.28 g= 980 cm/s^2 sqrt of g= 31.3 2 pi/ sqrt of g= .2 This is the same as in the earlier equation T= .2 * sqrt of L confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:35:40 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 --> L will be in cm if you use T= .20 * sqrt of L self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:37:32 `q006. If we wished to construct a pendulum with a period of exactly one second, how long would it have to be?
......!!!!!!!!...................................
RESPONSE --> 1s= period 1s= .2 * sqrt of L Divide both sides by .2 5=sqrt of L Square both sides 25cm=L confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:39:13 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 --> T= 2 pi/sqrt of g * sqrt of L T= 2 3.14 * sqrt (L/9.8m/s^2) Solve for L L =.26m self critique assessment: 2
.................................................
......!!!!!!!!...................................
10:45:59 `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= 300N/m m= 10 kg omega= sqrt of K/m omega= sqrt of 300N/m/10 kg= 5.5 rad/s T= 2 pi/5.5 rad/s= 1.14s 1/1.14s= .88 cycles/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:47:23 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 --> I left off a zero in the value of k. I used 300N/m. omega= 17.4rad/s T= 2 pi rad/ 17.4 rad/s= .36 s 1/.36s/cycle= 2.8 cycles /s self critique assessment: 2
.................................................
......!!!!!!!!...................................
11:05:30 `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.
......!!!!!!!!...................................
RESPONSE --> 60 s= 2 pi * sqrt of L/g 60s=6.28 * sqrt of L/9.8m/s^2 9.55s= sqrt of L/9.8m/s^2 Square both sides 91.24s^2= L/9.8m/s^2 Multiply by 9.8m/s^2 894.2m=L K= mg/L K= 55kg*9.8m/s^2/.248 m=2173N/m confidence assessment: 1
.................................................
......!!!!!!!!...................................
11:08:48 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 --> 45cycles=60s =1.33s/cycle 2 pi rad/1.33s= 4.7 rad/s omega= sqrt of k/m k=m*omega^2= 55 kg * (4.7 rad/s)^2=1200N/m self critique assessment: 2
.................................................