Energy and Simple Harmonic Motion - Part II.

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SUMMARY

The discussion focuses on calculating the speed of an object attached to a spring undergoing simple harmonic motion. The spring is compressed by 0.064 m and oscillates with an angular frequency of 11.5 rad/s. Using the formula for maximum velocity, V(max) = A(ω), the maximum speed is determined to be 0.736 m/s. The final speed at a stretch of 0.037 m is calculated to be 0.601 m/s using the conservation of energy principles and Hooke's Law.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Knowledge of angular frequency and its relation to simple harmonic motion
  • Familiarity with conservation of energy principles in mechanical systems
  • Ability to perform calculations involving square roots and algebraic manipulation
NEXT STEPS
  • Learn how to derive the spring constant using Hooke's Law
  • Study the relationship between angular frequency and frequency in oscillatory motion
  • Explore energy conservation in different mechanical systems
  • Investigate the effects of damping on simple harmonic motion
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for practical examples of energy conservation in spring systems.

sailordragonball
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A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall whle the other end is connected to a movable object. The spring and object are compressed by 0.064 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 11.5 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.037 m relative to its unstrained length?

... I tried using KEo + PEo + SPEo = KEf + PEf + SPEf ... but, I can't get anywhere ...

... I found out that the frequency is 18.05hz and it's relative time is .056 seconds - and that's where I'm stuck!
 
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Use Hooke's law and conservation of energy.
 
Do I set them equal to each other??

... I don't understand the .64 meter compression part ... does that equal "X" in Hooke's Law and in ?
 
Hint -- Can you find the value of the spring constant?

BTW -- you are right in that the compression WILL be X in hooke's Law.
 
I figured it out ...

A = 0.064m (given in the problem)
omega = 11.5 (rad/s) - given in the problem
x = 0.037m (given in the problem)V(max) = A(omega)
V(max) = 0.064(11.5)
V(max) = .736(m/s)

V = V(max) * [the square root of the quantity of (1- {[x^2]/[A^2]})]
V = .736 * [the square root of the quantity of (1- {[0.037^2]/[0.064^2]})]

V = 0.601 (m/s)
 

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