Finding amplitude of SHM with a theta vs time graph

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tdoyzz
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1. What is the amplitude of the pendulum's motion, in meters? (see attached graph)
Given: angular frequency=1.26rad/s



2. T=2pi(L/g)^.5; a=-w^2(x); s(t)=Acos(wt+phase); theta=s/L



3. I almost don't even know how to start.. The confusing part I believe is the graph. The max angle says its 1.. so I solve for L and got 6.19m, with the period being 5s. But I still keep getting wrong answers. Please help. Thanks
 
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tdoyzz said:
The max angle says its 1.. so I solve for L and got 6.19m,
That looks about right to me (possible rounding/precision issues though). :approve:
with the period being 5s.
Also seems reasonable. :approve:
But I still keep getting wrong answers. Please help. Thanks
The problem statement is asking for the amplitude of oscillation, not the length of the string. But since you know the length of the string (from above), and you know the max θ, how do you find the max amplitude?
 
Would the amplitude equal the max arc length?

A = s max = theta max/L = .16

This is wrong as well though..
 
tdoyzz said:
Would the amplitude equal the max arc length?

A = s max = theta max/L = .16

This is wrong as well though..
'Couple of things not quite right. You're dividing when you should be multiplying, and secondly, the arc length equation,
smax = θmaxL
(where smax is the arc length) only works if θmax is expressed in radians. :wink:
 
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Ah of course! Its always the simple things you miss.. Thanks