Simple Harmonics and also some pressure

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SUMMARY

The discussion focuses on solving problems related to simple harmonic motion and pressure forces on a door. The first problem involves determining the displacement at which kinetic and potential energies are equal for a simple harmonic oscillator with an amplitude of 0.1 m, using the formula x = A/√2. The second problem calculates the net force on a door with an area of 2 m², given a pressure difference of 0.01 atm, resulting in a net force of 4052 N, indicating difficulty in opening the door by hand.

PREREQUISITES
  • Understanding of simple harmonic motion and energy conservation
  • Knowledge of pressure calculations using the formula p = F/A
  • Familiarity with the concepts of kinetic and potential energy in oscillators
  • Basic grasp of atmospheric pressure and its effects on surfaces
NEXT STEPS
  • Study the principles of simple harmonic motion in detail, focusing on energy relationships
  • Learn how to apply the equations of motion for oscillators, specifically v = dx/dt
  • Investigate fluid mechanics concepts related to pressure differences and forces on surfaces
  • Explore practical applications of pressure calculations in engineering contexts
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as engineers interested in fluid dynamics and pressure-related calculations.

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Homework Statement


Hi

I have two problems i need to solve mainly 1, but i need to check if 2a and 2b is correct
here it is:

  1. A simple harmonics oscillator has an amplitude of 0,1m.At what displacement will its kinetic and potential energies be equal
  2. a) What is the force due to the atmosphere on one side of door of area 2m2?
    b) If the difference in pressure on the two sides of a closed door of area 2m2 is 0,01 atm, what is the net force on the door? Do you think you could open it by hand?

Homework Equations


in first problem i though of using x=A/√2 where A is the amplitude but i have no idea how to solve this in any way other.

In the two other problems i though of using p=F/A and then in last problem find the difference with F1-F2

The Attempt at a Solution



1) Though trying putting the x=A/√2 = 0,07 but I am not sure this is correct really since i just tried googling around for the answer and looking in the book.
Since our teacher give us this hint: v = dx/dt for x = Acoswt but idk how to use this.

2a) First i found F=P*A = 101300 Pa * 22 = 405200N

b) First i found the force F=P*A = 101300 Pa*0,99*2m2 = 401148 N

where the difference is F1-F2 = 405300-401148 = 4052N which would make it quite hard to try open the door
 
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