Calculating Amplitude in Simple Harmonic Motion

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SUMMARY

This discussion focuses on calculating amplitude in simple harmonic motion (SHM) through two specific problems. In the first problem, a mass of 3kg attached to a spring with a spring constant of 15 N/m is displaced by 0.1m, establishing that the amplitude of oscillation is indeed 0.1m. The second problem involves a 4kg mass on a 3m pendulum, where the maximum angle of swing can be determined using conservation of energy principles, despite initial velocity being horizontal at 0.4m/s.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with Hooke's Law and spring constants
  • Knowledge of pendulum dynamics and energy conservation
  • Proficiency in using trigonometric functions in physics equations
NEXT STEPS
  • Study the derivation and application of Hooke's Law in SHM
  • Learn about energy conservation principles in pendulum motion
  • Explore the mathematical relationships in SHM using equations of motion
  • Investigate the effects of damping on amplitude in oscillatory systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to amplitude in simple harmonic motion.

asteeves_
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I have encountered two separate review problems that have to do with finding a value for amplitude and I am really struggling with it.

1. Homework Statement

Question 1-
A mass of 3kg is free to move on a horizontal frictionless surface and attached to a spring of k=15 N/m. It is displaced from equilibrium by 0.1m to the right and released from rest. What is the amplitude of the oscillation ?

Question 2-
A simple pendulum consists of a 4kg mass attached to a 3m long rope. At t=0 it is moving to the right at 0.4m/s. What is the maximum angle to which it swings?

Homework Equations



(1) x=Acos(wt+θ)
(2) v=-wAsin(wt+θ)

The Attempt at a Solution


[/B]
Question 1-
I assumed the amplitude was equal to 0.1m from the question however I have a lot of trouble justifying it to myself. If this is intact the case I would really appreciate an explanation as to why, if not some insight on how to find A would be very helpful.

Question 2-
I attempted to solve for A using equation 2 and the information given regarding t=0 but ran into the issue of not knowing the value of θ, and have kind of hit a road block now.
 
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asteeves_ said:
I have encountered two separate review problems that have to do with finding a value for amplitude and I am really struggling with it.

1. Homework Statement

Question 1-
A mass of 3kg is free to move on a horizontal frictionless surface and attached to a spring of k=15 N/m. It is displaced from equilibrium by 0.1m to the right and released from rest. What is the amplitude of the oscillation ?

Question 2-
A simple pendulum consists of a 4kg mass attached to a 3m long rope. At t=0 it is moving to the right at 0.4m/s. What is the maximum angle to which it swings?

Homework Equations



(1) x=Acos(wt+θ)
(2) v=-wAsin(wt+θ)

The Attempt at a Solution


[/B]
Question 1-
I assumed the amplitude was equal to 0.1m from the question however I have a lot of trouble justifying it to myself. If this is intact the case I would really appreciate an explanation as to why, if not some insight on how to find A would be very helpful.
The mass is displaced by 0.1 m to the right from equilibrium, and then released at t=0. It means that the velocity is 0 at t=0. Substitute t=0, x=0.1 and v=0 into the equations (1) and (2). What do you get for θ and A?

asteeves_ said:
Question 2-
I attempted to solve for A using equation 2 and the information given regarding t=0 but ran into the issue of not knowing the value of θ, and have kind of hit a road block now.
The pendulum moves to the right at t=0, not to the right and up, so it moves horizontally at that instant. At what angle does it move horizontally?
(You can apply also conservation of energy to get the maximum angle it swings.)
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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