Simple Pendulum Motion & Time Period Formula with SHM Approximation

shadowrunner
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Homework Statement



The motion of a simple pendulum is only approximately simple harmonic. How small should the amplitude should be for the approximation to hold good?. Obtain the general expression for the time period of a simple pendulum. How much does the actual time period differ from the approximate time period when the amplitude is 15 degree?

Homework Equations




The Attempt at a Solution


On obtaining the period of oscillation using SHM, we approximate sin(a)=a in radians. I was thinking of keeping it as sin(a). Any help with the first question? The amplitude one.
 
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The approximation \sin\theta\approx\theta is derived via Taylor series expansion. To determine how accurate it is, you need to look at the Taylor series remainder.
 
\ T=2pi \sqrt{L/g}(1\theta^2/16+11\theta^4/3072+173\theta^6/737280+22931\theta^8/1321205760+...)

That was the equation I got for time period. But I can't approximate the smallest value of amplitude.
 
Last edited:
shadowrunner said:
\ T=2pi \sqrt{L/g}(1\theta^2/16+11\theta^4/3072+173\theta^6/737280+22931\theta^8/1321205760+...)

Surely you mean

T=2\pi\sqrt{\frac{L}{g}}\left(1+\frac{1}{16}\theta_0^2+\frac{11}{3072}\theta_0^4+\frac{173}{737280}\theta_0^6+\ldots\right)

where \theta_0 is the initial amplitude (angular displacement) of the pendulum...right?

That was the equation I got for time period. But I can't approximate the smallest value of amplitude.

Read the question again. You aren't asked to approximate the smallest value of amplitude.
 
"How small should the amplitude should be for the approximation to hold good?"
It is given in the question. I don't know how to do it.
 
The approximation they are referring to is

T=2\pi\sqrt{\frac{L}{g}}\left(1+\frac{1}{16}\theta _0^2+\frac{11}{3072}\theta_0^4+\frac{173}{737280}\theta_0^6+\ldots\right)\approx2\pi\sqrt{\frac{L}{g}}

...how large can you make \theta_0 before this is no longer a good approximation?
 
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