Simple Harmonic Motion Pendulum problem

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SUMMARY

The discussion centers on calculating the maximum speed of a simple pendulum with a length of 30.0 cm, released from an angle of 10.0 degrees. The relevant equations include the maximum speed formula v = ωA and the period T = 2π√(L/g). The correct amplitude A is determined to be 0.3 - 0.3cos(10°), leading to a maximum speed calculation of approximately 0.02606 m/s. However, the initial calculation was incorrect due to misinterpretation of amplitude.

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bobbricks
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Homework Statement


A simple pendulum of length =30.0cm is released from rest from an angle of θ=10.0∘ to the vertical.
Assuming that the pendulum undergoes simple harmonic motion, find its maximum speed.

Source: https://isaacphysics.org/questions/accuracy_shm_pendulum_num

Homework Equations


a) v=wA
b) T=2pi*root(L/g) --possibly?

The Attempt at a Solution


A=0.3-0.3cos10
w=2pi/T
so w=root(g/L) by subbing equation a) into b).
g=9.81 and L=0.3.

so v=root(9.81/0.3)*(0.3-0.3cos10)
=0.02606... ms^(-1)

But that isn't the correct answer?
 
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A is the amplitude of oscillation, not the length of the pendulum.
 
Is A=0.3-0.3cos10 ?
 
Bump
 
bobbricks said:
Is A=0.3-0.3cos10 ?

How did you come up with this number? How far away from the equilibrium is the pendulum when it is at an angle of 10 degrees?
 
Its maximum velocity will be at the lowest point. The total length is 0.3m. Its current vertical height distance from the top is 0.3cos10. So the difference is 0.3-0.3cos10 which is the maximum amplitude..?
 

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