how to find quality factor of a simple pendulum given its mass,period,amplitude?
85181
The real pendulum is not described by sines and cosines, or even damped sines and cosines, but matheiu functions. \n\nLd/dt(dA/dt)+g sin A=0\n\nwhere L is the length of pendulum, A is the angle of displacement, t is time, g is the acceleration of gravity.\n\nFor small A, this is approximated as\n\nLd/dt(dA/dt)+gA=0\n\nHowever, this still does not take a lot of other factors into account, that would produce a loss of energy. The quality factor is only relevant if energy is lost as the pendulum oscillates. (If there is no energy loss, you will have an infinite quality factor, since quality factor is inversely proportional to damping factor, and if there is no damping, then the Q must be infinite). You have to add frictional terms, and then solve or approximate the solution to the equation for small angular displacement, or do a brute force exact solution. \n\nThe second reference describes how to relate the parameters of the solution to your equation to the quality factor. \n\nA simple version would be to solve the equation\n\nLd/dt(dA/dt)+bdA/dt+gA=0\n\nwhere b is some sort of frictional factor, that can contain the effects of energy loss in the string, frictional force in the air, and so on. The references show how to relate this to the quality factor.
871463
The period is independent of the mass.\n\nT = 2pi*root(L/g)\n\nSurely this is derived in your text book.....?\n\nUse the the 17m and g = 9.8-------->  answer in seconds
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
840333
You need to consider conservation of angular momentum\n\nI0*w0 = If*wf\n\nI0 is the initial moment of inertia, If is the final moment of inertia, w0 is the initial angular speed, and wf is the final angular speed.\n\nI = Iss + Io\n\nwhere Iss is the (constant) moment of inertia of the student and the stool, and Io is the moment of inertia of the objects. This general relation applies to both the initial and final moments of inertia. Io can be computed from\n\nIo = 2*m*r^2\n\nwhere m is the mass of the object and r is the distance from the axis of rotation. The factor of 2 accounts for the two different objects. The first equation then becomes\n\n(Iss + 2*m*r0^2)*w0 = (Iss + 2*m*rf^2)*wf\n\nwhere r0 is the initial radius (1.0 m) and rf is the final radius (0.40 m). Solving for the final angular speed you have\n\nwf = (Iss + 2*m*r0^2)*w0/(Iss + 2*m*rf^2)\n\nUsing this expression it is easy to compute the final angular speed.
861187
0.4*(100kg*g+F*sin(x)) = F*cos(x). ignoring any rotational component
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
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.
956961
For an equation of the form y = a cos (bx + c) + d:\n\na is the amplitude\n2 pi / b is the period\nc is the phase shift\nd is the vertical shift.\n\nTherefore, the amplitude is 2, the period is 2 pi / 3, and the vertical shift is +4.
102434
Part (a): \n   Given three values (Mrock = 1.2 kg, Xrock = 5 cm, Xstand = 14 cm), and we need to solve for one unknown (Mstick).  To clarify, Mrock means mass of rock, and Xrock means position of the rock on the meter stick.\n   Since you're being tested on this concept, I'll assume you know the equation for torque: T = Force x Lever arm.  We will also use the concept of Static Equilibrium, which is a fancy term that means that nothing is moving.  Since this system is in static equilibrium, we can use the equation T(clockwise) = T(counterclockwise), using the stand as our pivot point (14 cm).\n   To find clockwise torque, we need to find the Lever arm from the center of mass to the pivot point.  The center of mass is located at 50 cm, and the pivot point is at 14 cm, so our lever arm is 50 cm - 14 cm = 36 cm.  T(clockwise) = Mstick * g * 36 cm  (It's okay to use units of cm in this problem because it cancels out in the end.)\n   If the rock is hung from a massless string, we can assume that the force on the meter stick from the string, which is the tension in the string, is equal to Mrock * g.\n   Counterclockwise torque: Lever arm = 14 cm - 5 cm = 9 cm.  Torque(counterclockwise) = Mrock * g * 9 cm (plug in numbers later to simplify calculation)\n   Torque(counterclockwise) = Torque(clockwise) ==>  Mrock * g * 9 cm = Mstick * g * 36 cm  ==> g's cancel, manipulate equation to get: Mstick = (9 cm / 36 cm) * 1.2 kg = 0.3 kg\n\nPart (b):\n   Since the stand is not causing the meter stick to accelerate upwards, it must be exerting a force upward equal to the force the meter stick exerts downward.  The total force from the meter stick is equal to (Mstick * g) + (Mrock * g) ===>  (Mstick + Mrock) * g ===>  1.5 kg * 9.8 m/s^2  ===> 14.7 N\n   I hope this helps (I'm a physics major...this should be a reliable answer)
792967
The period of a pendulum is given by \nT=2*Pi*Sqrt[lenght / g ]\nIn the equilibrium position all the energy is kinetic energy, so the total energy is give by:\nE=1/2*m*v^2\nIn the maximum angular position, all the kinetic energy is transformed into potential energy:\n1/2*m*v^2 = m*g*h\nThis way you can calculate the height and this you can convert into an angular position [just draw it, you'll see what to do]
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.
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
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
919525
Vibration:  ω = √(k/m) = √(100/4) = 5.0 rad/sec\n\nV at neutral point = Vo = Aω = 2*5.0 = 10 m/sec\n\nx-direction momentum: This momentum will be the same before and after dropping the 6 on the 4 →\n6*0 + 4*5 = (6+4)*v →  v = 20/10 = 2 m/s\n\nThe new KE = .5mv² = .5*10*2² = 20 Nm ; This will equal the new spring energy:\n\n20Nm = .5kA² → A = √(40/100) = .6324 m, so the change in amplitude is\na)  2-.6324 = 1.3676 m\n\nb) New period: T = 2π√(m/k) = 2π√(10/100) = 1.987 sec\n....Old period:  To = 2π√(4/100) = 1.2566 sec\n....Difference = .73 sec greater\n\nc)  KE1 = .5*4*10² = 200 Nm.  From (above) KE2 = 20 Nm\n....so, energy change is -180 Nm\n\nd) The energy loss is released as heat between the (momentarily) sliding surfaces of the 2 objects when they contact.
914856
OK as the ball is swung around in a horizontal circle the scale inside the tube is measuring the tension on the string. By knowing the time to complete one revolution  linear velocity can be found from v=r*omega(angular velocity). in the second part the v is given and needs to be converted into radians/second by using 3.7=.5*omega where omega=3.7/5=.74rad/s then using the mass radius and angular velocity you can find the tension in the string. With a= acceleration (into center) omega^2*r=.74^2*.5=.2738m/s^2and using f=ma the tension in the string will equal f which equals .2*.2738=.0548newtons Unless i have a silly math mistake somewhere.
375701
Try those...\n\nhttp://www.superbase.com/sb_scientific_current/example_pendulum.htm\n\nhttp://www.8886.co.uk/pendex1.htm
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)