Simple harmonic motion platform

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SUMMARY

The discussion focuses on a platform executing simple harmonic motion (SHM) with an amplitude of 5 cm and a frequency of 10/π vibrations per second. The key equations used include x=Acos(ωt-φ) and mg=-ma, with the angular frequency calculated as 20 rad/s. The primary questions addressed are the point at which a block leaves the platform and how far it will rise above the platform's highest point. The solution involves understanding the conditions under which the block detaches from the platform, specifically when the net force equals zero.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with angular frequency calculations
  • Knowledge of Newton's second law of motion
  • Ability to manipulate trigonometric functions in physics equations
NEXT STEPS
  • Study the derivation of angular frequency in SHM
  • Learn about the conditions for detachment in SHM systems
  • Explore the concept of maximum displacement in oscillatory motion
  • Investigate the role of phase angles in SHM equations
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for examples of simple harmonic motion applications.

Thatonetim194
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1. A platform is executing simple harmonic motion in a vertical direction with an amplitude of 5 cm and a frequency of 10/pi vibrations per second. a block is placed on the platform at the lowest point of its path.
a) at what point will the block leave the platform?
b)how far will the block reach above the highest point that the platform reaches

2. Homework Equations

x=Acos(ωt-phi)
mg=-ma(where the block leaves the platform)
g-(ω^2)Acos(ωt)=0
ω=2pif
m(g-(ω^2)Acos(ωt))=0

3. The Attempt at a Solution for part a

given the frequency i found the angular frequency to be 20rad/s

knowing x=-Acos(ωt) because it starts out at the lowest point at -A and also no phase angle to the equation.

to find when the mass leaves the platform -ma=mg, and taking the derivative twice of this equation a=(ω^2)Acos(ωt)

and i need to satisfy the initial condition of the acceleration so i get m(g-(ω^2)Acos(ωt))=0

but how do find when this happens?

part b I was not able to get to yet
 
Last edited:
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Thatonetim194 said:
1. A platform is executing simple harmonic motion in a vertical direction with an amplitude of 5 cm and a frequency of 10/pi vibrations per second. a block is placed on the platform at the lowest point of its path.
a) at what point will the block leave the platform?
b)how far will the block reach above the highest point that the platform reaches

2. Homework Equations

x=Acos(ωt-phi)
mg=-ma(where the block leaves the platform)
g-(ω^2)Acos(ωt)=0
ω=2pif
m(g-(ω^2)Acos(ωt))=0

3. The Attempt at a Solution for part a

given the frequency i found the angular frequency to be 20rad/s

knowing x=-Acos(ωt) because it starts out at the lowest point at -A and also no phase angle to the equation.

to find when the mass leaves the platform -ma=mg, and taking the derivative twice of this equation a=(ω^2)Acos(ωt)

and i need to satisfy the initial condition of the acceleration so i get m(g-(ω^2)Acos(ωt))=0

but how do find when this happens?

Don't they ask the position (they say "at what point"), not the time? Then you just need - Acos(ωt), no?
 
Just to let people know I actually found out how to do the problem sorry for posting this.
 

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