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    What is the reason that when the angle of release is increased (from say 10 to 80 degrees) the time period increases?

    are there any relevant formulae/relationships?

    and is this right:

    larger angle, higher centre of mass, greater gain in ke (loss of gpe), so greater gain in speed

    but with a greater angle a greater distance needs to be travelled

    and the rate of increase of distance with respect to theta is greater than the rate of increase of speed with respect to theta so the time period is longer


    and for plotting graph would you do T^2 against theta? or something involving logs or cosine of theta
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    due to fact that in real life air resistance would become much larger, if the angle of deviation is increased - it would therefore behave less in SHM way...though theoretically, the angle shouldn't make a difference in space, because there is very little air..its a vacuum.
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    air resistance would increase because the speed increases?
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    (Original post by MC REN)
    air resistance would increase because the speed increases?
    sort of - its because if the angle is bigger, the bob would travel through teh air faster, and i suppose you could assume that air resistance becomes proportional to the square of teh speed at high speeds, opposed to the speed, for small speeds
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    ...sigh...
    The reason it increases is due to the fact that the equation you were taught at ALevel is based on an approximation.
    The restoring force on a pendulum is F = -mg sinθ. However we, in the 'small angle approximation' use the Maclaren expansion of sinθ ≈ θ, to the first expansion term. Thus F ≈ -mgθ.
    However as θ increases sinθ ≠ θ, in fact sinθ < θ, thus |F| < |mgsinθ|. Thus at the large angles the restoring force is less than is required for pure SHM.
    QED
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    (Original post by Mehh)
    ...sigh...
    The reason it increases is due to the fact that the equation you were taught at ALevel is based on an approximation.
    The restoring force on a pendulum is F = -mg sinθ. However we, in the 'small angle approximation' use the Maclaren expansion of sinθ ≈ θ, to the first expansion term. Thus F ≈ -mgθ.
    However as θ increases sinθ ≠ θ, in fact sinθ < θ, thus |F| < |mgsinθ|. Thus at the large angles the restoring force is less than is required for pure SHM.
    QED
    One probem - you havn't defined your θ :cool: ...
    ....and to the thread starter, you dont need to know the maths behind it for GCSE, or A-level...just say its cos air resistance slows it down and therefore it would not prove to be a reliable model, etc...
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    but the problem is that having damping still results in simple harmonic motion, is that right?
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    As Mehh says it is not strictly due to air resistance. You could do the experiment in a vacuum and still observe deviations from SHM. It would be far better to talk about the use of the small angle approximation and how this no longer applies at large angles.
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    the equation is T = 2pi root(l/g)

    right?

    and the reasons are - air resistance, and more importantly - approximation of theta = sine theta becomes less and less accurate as size of theta increases

    cheers
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    BTW given linear air resistance (i.e. F = -bdθ/dt = -bω ). Where ω is angular velocity of the pendulum at time t.
    The angular velocity of a damped SHM system is ω = √(g/l - b²/4m²). Note the ω here is a different ω from above.
    Thus the frequency ν = (nu) = ω/2π = 1/2π x √(l/g - b²/4m²)
 
 
 
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