Oscillations of 1.5 kg Block on Spring: Frequency and Stretch

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SUMMARY

The discussion focuses on the oscillations of a 1.5 kg block attached to a vertical spring with a force constant of 300 N/m. The frequency of the resulting oscillations is calculated to be 2.3 Hz using the formula T = 2π√(m/k), where f = 1/T. Additionally, the maximum and minimum stretch of the spring during oscillations are determined to be 2.9 cm and 6.9 cm, respectively, based on the equilibrium position and amplitude of oscillation.

PREREQUISITES
  • Understanding of Hooke's Law (F = kx)
  • Knowledge of simple harmonic motion
  • Familiarity with oscillation frequency calculations
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the oscillation frequency formula T = 2π√(m/k)
  • Explore the concept of equilibrium position in spring systems
  • Learn about energy conservation in oscillatory motion
  • Investigate the effects of varying mass and spring constants on oscillation frequency
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Physics students, mechanical engineers, and anyone interested in understanding the dynamics of oscillatory systems and spring mechanics.

Dooga Blackrazor
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1. A block of mass 1.5 kg is attached to the end of a vertical spring of force constant k=300 N/m. After the block comes to rest, it is pulled down a distance of 2.0 cm and released.
(a) What is the frequency of the resulting oscillations?
(b) What are the maximum and minimum amounts of stretch of the spring during the oscillations of the block?

(a) 2.3 Hz
(b) 2.9 cm and 6.9 cm


I used the formula T = 2(pie)(squaroot of m/k) with f = 1 /T and found the first answer; however, I'm now sure how to calculate the answers for the second part of the question. F = k*x (spring constant times displacement, but I'm not sure how that factors into the two answers.

Thanks in advance.
 
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Dooga Blackrazor said:
I'm now sure how to calculate the answers for the second part of the question. F = k*x (spring constant times displacement, but I'm not sure how that factors into the two answers.
They are measuring the stretch of the spring from its unstretched position (before the mass is attached). First find out where the equilibrium position is when the mass is attached. Then realize that the mass will oscillate about that point with the given amplitude.
 
Thanks - I've got it now.
 

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