Finding Time in an Oscillation Problem

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SUMMARY

The discussion centers on calculating the oscillation period of a mass-spring system under different conditions. Initially, a mass oscillating on Earth with an amplitude of 2.4 cm takes 3.0 seconds for one complete oscillation. A second oscillator, with a mass six times heavier and a spring three times stiffer, is analyzed on a planet with a gravitational acceleration of 8.0 N/kg. Using the formulas for angular frequency and period, the new oscillation period can be derived based on the modified parameters.

PREREQUISITES
  • Understanding of harmonic motion and oscillation principles
  • Familiarity with Hooke's Law and spring constants
  • Knowledge of angular frequency and its relationship to oscillation period
  • Basic grasp of gravitational effects on oscillating systems
NEXT STEPS
  • Calculate the angular frequency for the new mass-spring system using w = sqrt(ks/m)
  • Determine the period T for the new oscillator using T = 2pi/w
  • Explore the effects of varying mass and spring stiffness on oscillation frequency
  • Investigate oscillation behavior in different gravitational fields
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to enhance their understanding of mass-spring systems.

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Homework Statement



Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 2.4 cm. You observe that it takes 3.0 s to make one round trip.

You construct another vertical oscillator with a mass 6 times as heavy and a spring 3 times as stiff. You take it to a planet where gplanet = 8.0 N/kg. You start it oscillating with amplitude 3.9 cm. How long does it take for the mass to make one round trip?

Homework Equations



x = Acos(wt)

w= sqrt(ks/m) = 2pi/T

T= 2pi/w

f = 1/T = w/2pi
 
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You know all formulae needed, what do you think?

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