What is the Required Frequency for Simple Harmonic Motion Testing?

AI Thread Summary
To determine the required frequency for simple harmonic motion testing of a hydraulic valve, the acceleration amplitude of 10g (98.1 m/s²) and a vibration amplitude of 2.0 cm are specified. The user initially considers using the formula T=1/(2π)√(l/g) for a simple pendulum, which is not applicable in this context. Instead, the correct approach involves using the relationship f=1/T, where T is the period derived from the appropriate equations for simple harmonic motion. The discussion emphasizes the need to clarify the definitions of vibrational amplitude and other relevant parameters. Understanding the correct equations and their application is crucial for solving the problem effectively.
Clairepie
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Homework Statement



I am confoosed, I think I need to find T first but not sure I am using the right equations, is vibrational amplitude both length & A?

A hydraulic valve is subjected to sinusoidal vibrations to test the component under conditions of simple harmonic motion.

The first test specifies an acceleration amplitude of 10g (= 98.1 m s−2) with a
vibration amplitude of 2.0 cm.
What frequency of vibration is required?

Homework Equations


I think these are the ones I should be using? If there are more I need to know!

T=1/(2pi) Sqrt (l/g)

f=1/T


The Attempt at a Solution


T=1/(2pi) Sqrt (l/g)
Where l=2.0 X 10-2 m
& g= 98.1 m s-2

Then f=1/T
 
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Clairepie said:

Homework Statement



I am confoosed, I think I need to find T first but not sure I am using the right equations, is vibrational amplitude both length & A?

A hydraulic valve is subjected to sinusoidal vibrations to test the component under conditions of simple harmonic motion.

The first test specifies an acceleration amplitude of 10g (= 98.1 m s−2) with a
vibration amplitude of 2.0 cm.
What frequency of vibration is required?

Homework Equations


I think these are the ones I should be using? If there are more I need to know!

T=1/(2pi) Sqrt (l/g)
This equation is for a simple pendulum: a mass m at the end of a string of length l. As there is no pendulum in this problem, the equation doesn't apply.
f=1/T


The Attempt at a Solution


T=1/(2pi) Sqrt (l/g)
Where l=2.0 X 10-2 m
& g= 98.1 m s-2

Then f=1/T
Hint: If you solve F=ma for simple-harmonic motion, you get a solution like x(t)=A cos(ωt+ϕ). You can refer to your notes and textbook if you're not sure what the various quantities in that equation stand for.
 
I think I have it, x(t)=A sin(ωt+ϕ) is the way forward, Thanks for clarifying that for me Vela *high five*

I read this, then took a shower & wrote down the algebra etc on the steamed up shower, I recommend this as the new "going for a walk"!
Clairepie

(If I came across a little daft, I can blame it on my medication & condition)
 
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