A Simple Harmonic Motion Problem

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SUMMARY

The discussion focuses on solving a simple harmonic motion problem involving a uniform rod of length L and mass M, pivoted at one end and subjected to two horizontal springs with force constants k1 and k2. The moment of inertia of the rod is calculated using the formula I = 1/3 ML². The solution for part (a) involves equating the moments from the springs and the rod's weight to find the angular frequency, leading to the period of small oscillations. For part (b), the maximum velocity of the endpoint P is derived using the relationship Vmax = ωθL.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with moment of inertia calculations
  • Knowledge of angular frequency and its relation to oscillations
  • Basic physics of springs and forces
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  • Study the derivation of the period of oscillation for different systems
  • Learn about the effects of damping on simple harmonic motion
  • Explore the concept of coupled oscillators and their applications
  • Investigate the role of energy conservation in oscillatory systems
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for examples of simple harmonic motion problems.

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Homework Statement



[PLAIN]http://img441.imageshack.us/img441/6951/physics.gif

As shown in the figure, one end (O) of a uniform rod of length L and mass M is pivoted to a wall. Two horizontal springs with force constants k1 and k2 are attached to the other end (P) of the rod. When the rod is aligned exactly along the vertical axis, the system is in equilibrium.

a) Find the period of small oscillations of the rod.
b) Assume that the initial angle the rod makes with y-axis is a small angle θ and the initial velocity 0. Find the velocity of the end point P of the rod as it passes through the equilibrium position.

Homework Equations



Moment of inertia of a rod relative to its end: I = 1/3 ML2

The Attempt at a Solution



I think I have a solution but it is too long and complex. I doubt it's correct. Any help appreciated.
 
Last edited by a moderator:
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Moment on the rod due to the spring is 2kl^2*theta. Equate it with I(alpha) then u will gte alpha and hence omega. From here u can proceed easily to find the period
 
That's what I did. But I added the moment from the mass of the rod too.

So k1L2θ + k2L2θ + MgLθ/2.

That's equal to Iα. (α being the second derivative of θ)

(k1L2θ + k2L2θ + MgLθ/2) / I is therefore w2.

That's part a. For part b, wmax = wθ and Vmax = wθL.

[PLAIN]http://img375.imageshack.us/img375/6378/captureoj.jpg

Are my answers correct? Thanks.
 
Last edited by a moderator:
yes u r correct...
 
Thanks a lot Swap.
 

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