Period of a pendulum?
827798
You haven't told us what point it's pivoting about.\nThis is a Physical Pendulum problem:\n\nT = 2π√(I/mgL)......for small angular oscillations....\n\nI = mR²  where R is the mean radius of the hoop\nL = the distance from the pivot to the center of mass of the hoop\n\nGOTTA have L............
824399
Period is independent of mass, so it doesn't matter how much the kid weighs. The period is\n\nT = 2 * pi *sqrt(L / g)\n\nwhere L = 5m, g = 9.81 and sqrt = square root
219785
f(x) = 2 sin[x - pi/3]\n\nI can see that x is not multiplied by another coefficient, so the wave graph is not altered by shrinking or expanding, and thus the period will not change, and the period is still 2pi.(it may be translated to the right by pi/3, but the size will not change)\n\n^_^
870396
Pendulum clock is clock based on a swinging weight. If you change the length of the swinging arm you change the time for each swing.\n\nSince wind resistance starts out as nothing and gets stronger, the slower (final speed) object reaches T.V.  first.\n\nThe problem with 1kg=2.2 pounds is pounds is force and kg is mass. But on the surface of the earth kg=kgf (kilogram force) so you should say 1 kgF=2.2lbs
151602
it would have no affect on a pendulum. \n\nstudies at the university of chico found that the up and down motion does not conflict in any way with the side to side motion of a pendulum. \n\nthe study also found that most of the research from the university of chico is conducted by a group of approximately 20 people ranging in age from 12 to 25.
908384
The frequency is 282 cycles per second (Hz), so one cycle will take 1/282 seconds. Period is the inverse (or reciprocal) of the frequency.
136500
Kepler's 3rd law says: Orbital period T = 2*pi*R^1.5/SQRT(GM)\nwhere R is the radius (in the case of a circular orbit), G is the universal gravitational constant and M is the mass of the central body.  Your question is (I think) why Period is proportional to R^1.5.  In orbit, centrifugal acceleration CA balances gravitational acceleration GA.  (Please, no centrifugal/centripetal arguments.  I'm a boor; I admit it!)  CA=w^2R (where w is omega, the orbital angular rate which = 2pi/T).  So CA = (2pi/T)^2*R.  From Newton we know GA = GM/R^2.  Put 'em together, \nCA = GA\nR*(2pi)^2/T^2 = GM/R^2  \nT^2 = R^3*(2pi)^2/GM\nT = 2*pi*R^1.5/SQRT(GM)
909370
Well, the equation for the time period of a pendulum is\n\nT = 2*pi*sqrt(l/g)\n\nWhere pi is the mathematical constant, 'l' is the length of the string, and 'g' is the acceleration of gravity.\n\nSo substitute your values, and you then have the equation\n\n0.80 = 2*pi*sqrt( l  /9.8 )\n\nYou could solve for the length and plug that back into the equation.\n\nT = 2*pi*sqrt( l / (.55*9.8) )\n\nSubstitute the length in for 'l' and solve for T
940810
Rotation period = 1 day\nOrbital period = 1 year
970854
Use Kepler's law, which states that the orbital period squared is proportional to the cube of the orbital radius.  See http://en.wikipedia.org/wiki/Kepler%27s_law
572734
Frequency is measured in Hertz, or inverse seconds.  So take the inverse of 6, which would be 1/6, which is your frequency.  Tada!
725972
You mean the graph of the function?  The second term is a sine wave with frequency 2x, meaning it goes through a full cycle in pi radians instead of 2*pi radians.  The max amplitude of the second term is +/- 1/2, since the max of the sine is +/- 1.  That's the second term.  Now, offset that by 4 units down at every point.  Got it?  Okay, your turn.
500385
Depends a bit on what grade you're in.  I'm gonna guess that since this assignment is about measurement in experiments, you're at a pretty basic level in science.\n\nA pendulum is a great thing for doing experiments involving measuring.  All you need is two or three small weights of different masses, a piece of string to tie them to, and a fixed point from which to hang it and let it swing.  With this you can do experiments measuring distance, time and mass.  Does the pendulum's period (the time it takes to swing back and forth once) change as the length of the swing shortens?  Does it change if you change the mass of the weight?  How about the length of the string?  Or, measure how long it takes to go from one swing length (say, 6 inches) to almost standing still (about 1/4 inch for the cutoff).  Does this change with the length of the string, or the mass of the weight?\n\nYou could have a whole table of measurements of the different factors (length of swing, length of string, mass of weight) and determine wich factors are constant and which are variable.\n\n\nA similar experiment is the one where Galileo tested to see if a heavier object actually falls faster than a lighter one.  The old story is that he dropped balls of different weights off the Tower of Pisa.  But what he actually did, is he took balls that were the same size but different weights and rolled them down a ramp, and timed them to see how fast they rolled down.  Can you figure out why he did it this way, and didn't just drop weights from a height?\n\nGood luck!
742680
You said that your timing equipment has an accuracy of +- 33 milliseconds. In statistics, I can assume that the +- 33 milliseconds is a 95% confidence interval of the mean period of the pendulum. So, since, the critical value in a normal distribution for a 95% confidence interval is 1.96, therefore, your standard deviation is about 0.0168367 s. Working shown below,\n\n1.96 * stddev = 0.033\nstddev = 0.033/1.96 = 0.0168367\n\nSo, using the Central Limit Theorem, the standard deviation of the estimated mean is stddev/sqrt(n), where n is the number of measurements and sqrt is the square root.\n\nSo the accuracy (95% confidence interval) of the period would be\n1.96 * stddev/sqrt(n)\n= 1.96 * 0.0168367 / sqrt(n)\n= 0.033 / sqrt(n)\n\nSo as you can see the accuracy drops by a factor of 1/sqrt(n)\n\nSo, if you do 2 measurements, you get an accuracy of plus minus 0.033/sqrt(2) = 0.023 s, ie plus minus 23 milliseconds\n\nIf you do 3 measuremenets, you  get an accuracy of plus minus 0.033/sqrt(3) = 0.019 s, ie plus minus 19 milliseconds
190007
There is no oscillation as conical oscillation. But there is conical pendulum.\n\nA small bob attached to one end of a string, the other end being fixed, is made to move in a cicle in a horizontal plane with uniform speed.\n\nThe string sweeps over a surface of a cone. Therefore the pendulum is called conical pendulum.\n\nThe period of rotation depends on the height of the pendulum.
736992
http://en.wikipedia.org/wiki/Pendulum
761664
Pulsar's period, T = 0.033 sec and is increasing at a rate of 1.26 x 10^-5 sec per year\n\n(a) Angular acceleration, α = ∆ω/∆t\n\nwhere ω = 2π/T = 6.283/0.033 = 190.39 rad/sec\n\nSince the period increases by 1.26 x 10^-5 sec per year, then\n\n∆T = 1.26 x 10^-5 sec => T' = T + ∆T = 0.0330126 sec\n\nω' = 2π/T' = 6.283/0.0330126 = 190.32 rad/sec\n\nHence = ∆ω = ω' - ω = -0.07267\n\nNow, α = ∆ω/∆t = -0.07267/31536000 = 2.3 x 10^-9 rad/sec^2\n\n(b) ω' = ω + αt, set  ω' = 0, solve for t:\n\nt = -ω/α = (190.39)/(2.3 x 10^-9) = 82,624,112,123.83 sec\n\nor 2,620 years\n\n(c) ω = ω(0) + αt, set t = (2006-1054)x365x24x60x60 sec\n\nω(0) = ω - αt = 190.39 - (-(2.3 x 10^-9)(30022272000))\n\nω(0) = 190.39 + 69.05 = 259.44 rad/sec or its initial period\n\nT(0) = 0.024 sec