What is the frequency of the oscillation?

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SUMMARY

The discussion focuses on calculating the frequency of oscillation for a mass-spring system, where an object is attached to a massless spring. The frequency is derived using the formula f = ω/(2π), where ω = √(k/m). The participants also explore the effects of adding mass to the system, which results in a frequency reduction to half the original value. Additionally, they discuss the application of Newton's 2nd Law to derive the equation of motion for the system, leading to a second-order differential equation.

PREREQUISITES
  • Understanding of harmonic motion and oscillation principles
  • Familiarity with Newton's 2nd Law of motion
  • Knowledge of differential equations and their solutions
  • Basic concepts of spring constants and mass-spring systems
NEXT STEPS
  • Research the derivation of the mass-spring system frequency formula
  • Learn about the effects of mass on oscillation frequency in spring systems
  • Study the application of differential equations in mechanical systems
  • Explore the relationship between gravitational force and spring constant in oscillatory motion
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Students and professionals in physics, mechanical engineering, and anyone studying dynamics and oscillatory systems will benefit from this discussion.

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A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially
held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and
oscillates up and down, with its lowest position being 10 cm below yi.
(a) What is the frequency of the oscillation?
(b) What is the speed of the object when it is 8.0 cm below the
initial position?
(c) An object of mass 300 g is attached to the first object, after which the system oscillates
with half the original frequency. What is the mass of the first object?
(d) Relative to yi, where is the new
equilibrium (rest) position with both objects attached to the spring?
 
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Apply Newton's 2nd Law to get the equation of motion for this system.

\sum_{i=1}^{n} \vec{F}_{i} = m \vec{a}

mg - kx = m \ddot{x}

g - \frac{k}{m}x = \ddot{x}

this gives a second order differential equation with the following solution:

x = A \sin (\omega t - \phi) + B
 
For finding the frequency, I did mg=kx, and k=mg/x. since f=1/2pi*sqrt(k/m) I substitute mg/x with k. But it was wrong.
 
Last edited:
the frequency is:

<br /> f = \frac{\omega}{2\pi}<br />

and

<br /> \omega = \sqrt{\frac{k}{m}}<br />
 
Last edited by a moderator:
Hmm.. Did they tell you what the spring constant is?
 
no. :frown:
 
Actually, if it has the spring got a gravitational acceleration then you can say that.

<br /> g = \omega ^2x<br />

But i am still not sure about this.
 
If you have omega then:

for b) you can let x(t) = 8 and find t.
then take the derivative of x(t) and find the velocity.

I think that should work.
 
how did you come up with the equation g=w^2*x?
 
  • #10
well - if u take the second derivative of x(t)
 

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