Period of a mass oscillating on a spring

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SUMMARY

The period of a mass oscillating on a Hookean spring is defined by the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. The discussion confirms that both integration and direct application of Newton's second law lead to the same result for the period. The alternative method presented simplifies the derivation by using the relationship between acceleration and displacement, expressed as a = -(k/m)x, which aligns with the standard form of simple harmonic motion.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Basic knowledge of differential equations
  • Familiarity with concepts of simple harmonic motion
  • Ability to perform integration and manipulate algebraic expressions
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  • Study the derivation of simple harmonic motion equations
  • Learn about the energy conservation in oscillatory systems
  • Explore the effects of damping on oscillations
  • Investigate the relationship between mass, spring constant, and frequency
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Homework Statement



The period of a mass m oscillating on a hookian spring of negligible mass.

Homework Equations



F = -kx

The Attempt at a Solution



I'm using x in place of ∆x for convienance.

F = -kx

dv/dt = -kx/m

∫vdv = -k/m∫xdx

v^2 = C - k(x^2)/m: Placing the restraint that if v = 0, x = xi

v^2 = k(xi^2)/m - k(x^2)/m

v = sqrt(k(xi^2)/m - k(x^2)/m)

t = sqrt(m/k)∫dx/sqrt(xi^2 - x^2)

t = arcsin(x/xi)sqrt(m/k). Therefore, if we allow ∆x = 0

t = arcsin(0)sqrt(m/k). Hence, if we want the time for one period:

t = 2pisqrt(m/k)

Is this correct? Thanks!
 
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Yes that is correct.
But if you didn't want to integrate you could have just put F=ma=-kx
such that a= -(k/m)x
which is in the form a=-(w^2)x meaning that w=sqrt(k/m)

and the period T=2pi/w
 
Thanks! That's a much simpler derivation.
 

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