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Basic Pendulum question
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
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
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
...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
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
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
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
.......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...
but the problem is that having damping still results in simple harmonic motion, is that right?
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)
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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