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Thread: stuck

  1. #1 stuck 
    Forum Freshman
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    Jan 2006
    how does the length of the pendulum and the mass of the bob affect the period of the pendulum?? whats the logic behind??

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  3. #2  
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    Sep 2005
    (someone correct me if I'm wrong)

    For length :

    As you increase the length of the string, the distance the bob has to travel to complete an oscillation increases and thus the period increases.

    For mass :

    Since no energy is lost, kinetic energy = potential energy

    1/2 mv^2 = mgh ------> v = square root (2gh)

    So mass actually doesn't affect the speed and thus doesn't affect the period.

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  4. #3  
    JS is offline
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    Jan 2006
    The simplest way to derive the equations of motion for a mathematical pendulum is by considering the energy equation: mgh + ½mv^2 = E_0 at all times. Using the constraints imposed by the pendulum, we may rewrite this as mgl(1-cos(\theta)) + ½ml^2*\theta-dot^2 = E_0

    Since energy is conserved, E_0 is constant with time. By taking the differential with respect to time on both sides of the equation we arrive at mgl*sin(\theta)*\theta-dot + ml^2*\theta-doubledot*\theta-dot = 0

    Since \theta-dot =/= 0 (except at the ends of the motion, which are trivially easy to solve anyway), we may reduce this to \theta-doubledot = - g/l*sin(\theta). The Taylor series of sin(x) is x - 1/6x^3 + higher orders.

    So as long as the angle is so small that \theta >> 1/6\theta^3 we may substitute \theta for sin(\theta), and arrive at the differential equation \theta-doubledot = g/l*\theta This equation can be solved and the solution is \theta = A*cos(sqrt(g/l)*t + \phi)

    A and \phi are determined by the constraints of the problem. This is a one-dimensional harmonic motion with a period of 2\pi*sqrt(l/g)

    Does that answer your question?
    And God said: Div(E) = \rho / \epsilon_0, Curl(E) = - \partial(B,t), Div(B) = 0, Curl(B) = \mu_0 (j + \epsilon_0 \partial(E,t)). And there was light.


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