A Simple Harmonic Motion Problem

In summary, the conversation discusses a problem involving a uniform rod attached to a wall with two horizontal springs. The system is in equilibrium when the rod is aligned along the vertical axis. The problem asks for the period of small oscillations of the rod and the velocity of the end point as it passes through the equilibrium position. The solution involves calculating the moment on the rod from the springs and the mass, equating it to the moment of inertia, and solving for the angular velocity. The final answers are correct.
  • #1
Faux Carnival
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Homework Statement



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

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.
 
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  • #2
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
 
  • #3
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 [Broken]

Are my answers correct? Thanks.
 
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  • #4
yes u r correct...
 
  • #5
Thanks a lot Swap.
 

What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion where a system oscillates back and forth around a stable equilibrium position. This motion is characterized by a restoring force that is directly proportional to the displacement from the equilibrium point and acts in the opposite direction of the displacement.

What are the factors that affect Simple Harmonic Motion?

The factors that affect Simple Harmonic Motion are the mass of the object, the spring constant, and the amplitude of the oscillation. The period of SHM is also affected by the length of the pendulum (for a pendulum system) or the length of the spring (for a mass-spring system).

What is the equation for Simple Harmonic Motion?

The equation for Simple Harmonic Motion is x = A sin(ωt), where x is the displacement from the equilibrium position, A is the amplitude, ω is the angular frequency (ω = 2πf), and t is the time.

How do you calculate the period of Simple Harmonic Motion?

The period of Simple Harmonic Motion can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant. This equation assumes a mass-spring system with no damping.

What is the difference between Simple Harmonic Motion and Damped Harmonic Motion?

The main difference between Simple Harmonic Motion and Damped Harmonic Motion is the presence of a damping force. In damped harmonic motion, the amplitude of the oscillation decreases over time due to the presence of a damping force, while in simple harmonic motion, the amplitude remains constant.

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