Simple harmonic motion in a string

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SUMMARY

The discussion focuses on calculating the time it takes for a particle of mass 0.4 kg, attached to a light elastic string with a natural length of 50 cm and a modulus of 20 N, to return to its original position after being released from a distance of 60 cm. The tension in the string is defined by the equation T = 40x, leading to the acceleration equation a = -100x. The particle undergoes simple harmonic motion, completing a quarter cycle to return to the string's natural length before moving unimpeded to point O.

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  • Understanding of simple harmonic motion principles
  • Familiarity with Hooke's Law and elastic potential energy
  • Knowledge of differential equations related to motion
  • Basic physics concepts of mass, tension, and acceleration
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  • Study the principles of simple harmonic motion in elastic materials
  • Learn about Hooke's Law and its applications in real-world scenarios
  • Explore differential equations for modeling motion in physics
  • Investigate the differences between elastic strings and springs in terms of motion
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to simple harmonic motion in elastic materials.

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Homework Statement


Hi
I need some clarification on this question
A light elastic string of natural length 50 cm and modulus 20 N has one end fastened to O on a smooth horizontal table. The other end has a particle of mass 0.4 kg attached to it and the particle is released from rest at a distance 60 cm from O. Find the time it takes to reach O.


Homework Equations





The Attempt at a Solution


While the string is stretched
T=kx/l
=20x/0.5
=40x
40x=-.4a
So,a = -100x

w=10
T=2pi/10 =pi/5

I get stuck from here. Can anyone help please? Thanks
 
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Think about how the mass moves after it's released. If the string were a spring, the mass would take a quarter of a cycle to return to its natural length, another 1/4 cycle to get to its farthest point, and 1/2 cycle to come back.

However, this is a string, which means it can't resist any compression. The mass would take 1/4 cycle for the string to return to its natural length. After that, the mass moves unimpeded all the way to O and beyond.
 
Thanks a lot guys! Sorry for late reply.
 

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