Period of a simple harmonic oscillator

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SUMMARY

The discussion centers on calculating the period of oscillation for a simple harmonic oscillator, specifically a skydiver with a mass of 90.0 kg and a second skydiver of 60.0 kg. The initial period is given as 1.50 seconds, and the new period of oscillation when both skydivers are considered is determined to be 1.94 seconds. The relevant formula used is T = 2π√(m/k), which relates the period T to the mass m and spring constant k of the system.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with the formula T = 2π√(m/k)
  • Basic knowledge of mass and spring constant relationships
  • Ability to perform calculations involving square roots and trigonometric functions
NEXT STEPS
  • Study the derivation of the simple harmonic oscillator formula T = 2π√(m/k)
  • Learn about the effects of mass on the period of oscillation in different systems
  • Explore the concept of damping in oscillatory systems
  • Investigate the role of the spring constant k in various physical contexts
USEFUL FOR

Students preparing for physics exams, particularly those focusing on mechanics and oscillatory motion, as well as educators teaching concepts related to simple harmonic oscillators.

spraymonkey32
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Hi I'm having problems with solving this question:
a 90.0 kg skydiver hanging from a parachute bounces up and down with a period of 1.50 seconds. What is the new period of oscillation when a second skydiver, whose mass is 60.0 kg, hangs from the legs first?

the answer is 1.94 seconds


Homework Equations


I think it has something to do with the the simple harmonic oscillator:
T = 2 (pi) sqrt(m/k)
I also tried the simple harmonic equation involving sin:
y=sinAwt=Asin sqrt(k/m)t

but i keep on getting 0.129 seconds.


I attempted this several times and is stuck on how to do it. I am studying for exams and there are several questions similar to it. Can someone help me please! :)
 
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Can you show your calculations for the T = 2π... method?
 

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