In Part A of this investigation, how will you detrmine the period of the pendulum?
141587
You need to read the experimental procedure to see how they want you to do it.  If it doesn't specify, presumably you will need to use a stopwatch.  (The period is the time it takes for one complete swing.)  Time ten swings with a stopwatch and divide the answer by ten.  Do that again to convince yourself that it doesn't matter where you start the pendulum, the period remains the same.
753893
The period of a simple pendulum can be expressed as\n\nT = 2*pi*sqrt(L/g)\n\nwhere L is the length of the pendulum and g is the acceleration due to gravity (on whatever planet you happen to be on).
988946
Follow the previous answers to get the period. Once you obtain that, your equation will be y = sin(2pi*x/P), where P is the period. If you think about the regular sin curve, y = sin(x), this works out because it has a period of 2pi so you can see the cancellation.
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
105862
I don't see how they can do away with them. It is a fact that the time is not running exactly to the second every year. If they did away with them they would eventually have to adjust something or atomic time would be different then local time.
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.
638729
The grandfather clock works on a mechanical system of gears, spring, levers, catches and a pendulum for keeping time. Energy is stored in a wind-up spring or a weight driven axis.\n\nThe basic time keeping is achieved by a pendulum that counts time in an accurate fashion within seconds per month. This uses the principle of Periodic motion and may be adjusted to be slightly faster or slowly by shifting the pendulum ball's position slightly.\n\nThe pendulum counter then controls the movements of a system of mechanical levers, gears & wheels that are calculated to drive the second, minute and hour hands in the desired rotation ratios so that they will show the correct time.\n\nUpon the hour, the system will activate a chime via the spring driven hammers that strike tubular bells or other sounding vibrating elements. A series of activation may be composed to produce the desired melody. Some system will also provide the date for the month. In its peak, many other noviety displays such as birds, toys, animals, etc would be used to announce the top of the hour much to the delight of children and adults alike.
783790
F=-kx and F=ma\n\nfind mass from period     T = 2pi(m/k)^(1/2)\n\nhmmmm,  you seem to be missing k\n\nfor fixed ratio m/k the period of 0.51s and amplitude 4 cm is not enough to determine the mass (and therefore acceleration).
854810
Since benabean took the absolute difference in the periods, the value of K IS relevant; in fact, crucial to the solution.  My solution:\nTake the ratio of the 2 periods: T2/T1 = √(g1/g2) = 1.00095426\nMultiply this by the seconds in a day (86400)→ 86482.4\n\nSo, a difference of 82.4 sec lost per day.
721327
Let's say the pendulum starts on the left. It passes through the eq. point, then moves to the right, then passes the eq. point again, then back to the left. If it took 1 second to move from the eq. point to the right and back again, then it took a 1/2 second to get there from the left and to swing back to the left. Therefore, the period is 2s.
115436
With a Magnetometer. See below for more detail.
757410
This equation only works for small angles theta < 10 degrees\n\nPeriod = 2*pi * sqrt (length / g)\n\nThe period in an undamped simple pendulum does not depend on mass.\n\nThere is a more in depth formula for larger angles on hyperphysics.
312323
Take the difference between the x values.\nThe period is the horizontal stretch, which is also called the wavelength.  They're all one and the same, from one crest of a wave (or any other part) to the next.\nI'm talking about sine and cosine waves here.
698608
Have you been given the standard formulas?\nx = a cos (nt + c),  frequency = n, period = 2*pi/n, phase = c (Not sure of that -- it may be -c.  Look it up.) This gives you the answers to (d), (e) and (f) immediately.  [No, I'm not doing it for you: You can substitute values I'm sure.]\n\nx is the displacement, so just sub t = 2 in this equation.  If you need touse your calculator (you shouldn't!), make sure it's set in radians.\n\nVelocity is dx/dt.  You can do that I hope, and get\n\nv = - 18*pi*sin(3*pi*t + pi/3)  Notice I omit units.  I prefer to use equations relating to numbers only (though I know units are sometimes useful especially in a Physics context) and then state the correct units in presenting the answer -- which you are going to do!\nSub t = 2 to get v in m/s\n\nAcceleration is dv/dt, so differentiate and then sub t = 2\n\nAll the best
799062
All you have to do is write an equation for the displacement of the springs as a function of x.  When the the particle is pulled to any x, the springs form 2 right triangles with the x and y axes:\n\n&#92;\n &#92;\n  @\n /\n/\n\nso lets call the length of the stretched springs h and define the displacement d = h-L:\nh^2 = x^2 + y^2\ny = L\nh^2 = x^2 + L^2\nh = sqrt(x^2 + L^2)\nd+L = sqrt(x^2 + L^2)\nd = sqrt(x^2 + L^2) - L\n\npotential in a spring is equal to 1/2 times the constant times the displacement squared.  So for one spring:\n\nPE = 0.5*k*d^2\nand for both springs, double it:\nPE = k*d^2\nPE = k*(sqrt(x^2 + L^2) - L)^2\n\nTo find the forces you have to do some more trig.  The total force of one spring equals k times the displacement.  In the y direction, you know that the net forces will always be 0 because the two springs cancel out.  In the x direction, the x components will be added.  x components are found be multiplying the magnitude times cos(theta) where theta is the angle formed. So:\n\nF_y = 0\nF_x = k*d*cos(theta) + k*d*cos(theta)\nF_x = 2*k*d*cos(theta)\nF_x = 2*k*(sqrt(x^2 + L^2) - L)*cos(theta)\n\ncos(theta) = adjacent/hypotenuse\ncos(theta) = x/h\ncos(theta) = x/(d+L)\ncos(theta) = x/(sqrt(x^2 + L^2) - L +L))\ncos(theta) = x/sqrt(x^2 + L^2)\n\nF_x = 2*k*(sqrt(x^2 + L^2) - L)*x/sqrt(x^2 + L^2)
697951
find out what happens to Y (what it approaches) as:\nx goes to infinity\nx goes to negative infinity\nthen find out what happens to X as:\ny goes to infinity\ny goes to negative infinity\n\nalso find the intercepts (if they exist)
225581
see if this helps.....\n\n\n\nAnalysis of a simple gravity pendulum\n \nFigure 2. Force diagram of a simple gravity pendulum.To begin, we shall make three assumptions about the simple pendulum\n\nThe rod/string/cable on which the bob is swinging is massless and always remains taut; \nThe bob is a point mass; \nMotion occurs in a 2 dimensional plane, i.e. pendulum does not swing in and out of the page. \nConsider Figure 2. The blue arrow is the gravitational force acting on the bob, violet arrows are that same force resolved into components parallel and perpendicular to the bob's instantaneous motion, the motion along the red axis, which is always perpendicular to the cable/rod. Newton's second law\n\n \nwhere F is the force acting on mass m, causing it to accelerate at a meters per second2. Because the bob is constrained to move on the green circular arc, there is no need to consider any force other than the one responsible for instantaneous acceleration parallel to the known path, the short violet arrow in our case\n\n \n \nLinear acceleration a along the red axis can be related to the change in angle θ by the arc length formula\n\n \n \n \nThis, however, is not the acceleration we seek because the gravitational force on the bob causes a decrease in angle θ. One accounts for that by placing a negative sign in front of a, thus:\n\n \n \n \nFigure 3. Trigonometry of a simple gravity pendulum.This is the differential equation which, when solved for θ(t), will yield the motion of the pendulum. It can also be obtained via the conservation of mechanical energy principle: any given object which fell a vertical distance h starting from rest, would have acquired kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy. Change in potential energy is given by\n\n \nchange in kinetic energy (body started from rest) is given by\n\n \nSince no energy is lost, those two must be equal\n\n \n \nUsing the arc length formula above, this equation can be re-written in favor of \n\n \nbut what is h? It is the vertical distance the pendulum fell. Consider Figure 3. If the pendulum starts its swing from some initial angle θ0, then y0, the vertical distance from the screw, is given by\n\n \nsimilarly, for y1, we have\n\n \nthen h is the difference of the two\n\n \nsubstituting this into the equation for  gives\n\n \nThis equation is known as the first integral of motion, it gives the velocity in terms of the location and includes an integration constant related to the initial displacement (θ0). We can differentiate, by applying the chain rule, with respect to time to get the acceleration\n\n \n \n \nsame as obtained through force analysis.\n\n[edit]\nSmall angle approximation\nThe problem with the equations developed in the previous section is that they are unintegrable. To shed some light on the behavior of the pendulum we shall make another approximation. Namely, we restrict the motion of the pendulum to a relatively small amplitude, that is, relatively small θ. How small? Small enough that the following approximation is true within some desirable tolerance\n\n \nif and only if\n\n \nSubstituting this approximation into (1) yields\n\n \nSolution to this equation is a well-known, and quite expected, oscillatory function\n\n \nwhere θ0 is the semi-amplitude of the oscillation, that is the maximum angle between the rod of the pendulum and the vertical.\n\nThe term  is a pulsation, which is equal to ,\n\nwhere T0 is the period of a complete oscillation (outward and return).\n\nSince ,\n\nthe period of a complete oscillation can be easily found, and we have obtained Huygens's law:\n\n \n \n[edit]\nArbitrary-amplitude period\nFor amplitudes beyond the small angle approximation, one can compute the exact period by inverting equation (2)\n\n \nand integrating over one complete cycle,\n\n \nor twice the half-cycle\n\n \nor 4 times the quarter-cycle\n\n \nwhich