Approaching Simple Harmonic Motion Amplitude Calculation

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SUMMARY

The discussion centers on calculating the amplitude of motion for a block on a piston undergoing simple harmonic motion with a period of 1.18 seconds. The user, Mike, correctly identifies that the condition for separation is when the inertial force (ma) is less than the gravitational force (mg). By substituting the acceleration formula, a = (2π × frequency)² × A, into the inequality ma < mg, Mike can solve for the maximum amplitude A, confirming that his approach is valid.

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  • Understanding of simple harmonic motion principles
  • Familiarity with Newton's second law (F = ma)
  • Knowledge of oscillation frequency and its relation to period
  • Ability to manipulate algebraic equations for solving inequalities
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  • Study the relationship between frequency and period in simple harmonic motion
  • Learn how to derive acceleration in oscillatory systems
  • Explore the conditions for separation in oscillatory motion
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Students and educators in physics, mechanical engineers, and anyone interested in the dynamics of oscillatory systems will benefit from this discussion.

sopi
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Hello, I was wondering if I am approaching the following problem correctly:

"A block is on a piston that is moving vertically with simple harmonic motion.
At what amplitude of motion will the block and the piston separate if the period of the piston is 1.18 s?"

I am currently doing the following:

We know F = ma must be less than F= mg. We also know:
a = acceleration = (2pi x frequency)^2 times A(max amplitude).

So then we sub 'a' into 'ma < mg' and solve for amplitude A. Is this correct?

Thanks,
Mike
 
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Yes, this is a correct approach to solve the problem. You can use the values given to calculate the acceleration and then solve for the amplitude.
 

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