Rods & Springs Oscillations: Frequency Calculation

[thejinx0r]

Homework Statement

A rod of length l and mass m, pivoted at one end, is held by a spring at its midpoint and a spring at its far end,on both pulling in opposite directions. The springs have constant k and at equilibrium, their pull is perpendicular to the rod. Find the frequency of small oscillations.

Homework Equations

[tex] \tau = R \times F[/tex]
[tex]F=-kx[/tex]

The Attempt at a Solution

Consider the origin to be the pivot point.

So :
[tex]\tau = R \times F [/tex]
[tex] =\frac{l}{2}\cdot (kx)cos(\theta) - l \cdot kx + l/2 \cdot cos(\theta) + mg \cdot sin(\theta)[/tex]
I am a little unsure here.
First theta is the angle between the vertical and the rod.
I am unsure because clearly when the spring in the middle is stretched by x, the one at the far end must be compressed twice as much, but I'm not sure if the factor of 2 is taken care of by the "R" for the torque or if I would have to apply it twice, once for the "R" and once for "x" from kx.

Anyways:
[tex] \tau = I \alpha [/tex]
[tex] = \frac{1}{3} M \cdot l^2 \alpha[/tex]
[tex] = \frac {1}{3} M \cdot l^2 \frac{d}{d\theta}(0.5 \omega^2)[/tex]
[tex]\tau = \frac {1}{6} M \cdot l \frac{d}{d\theta}( \omega^2) = -kxcos(\theta) + l/2mgsin(\theta)[/tex]

After that, I'm not sure what to do :S

I would integrate with respect to theta to get omega, but x is a function of theta?
And suppose that x was not, then omega^2 would be negative for certain values k and M.

 

[physics girl phd]

I didn't check your earlier work, but based on your last line: Maybe do Taylor's expansions of cos and sin? You'll have to play with how far to keep higher orders, so you get something solvable yet not trivial.

 

[thomate1]

I would approach this problem by first identifying the key variables and equations involved. From the problem statement, we can see that the key variables are the length of the rod (l), its mass (m), and the spring constant (k). The key equation to use would be the equation for the period of a simple harmonic oscillator:

T = 2π √(m/k)

To apply this equation to the given scenario, we first need to determine the effective mass of the system. This would involve taking into account the mass of the rod and the mass of the springs, as well as the way in which they are connected. Once the effective mass is determined, we can plug it into the equation to calculate the period (T) of the oscillations.

To find the effective mass, we would need to consider the rotational inertia of the system, which would involve integrating the mass of the rod and the springs along the length of the rod. This would also take into account the fact that the springs are pulling in opposite directions and the rod is pivoted at one end.

After finding the effective mass, we can then plug it into the equation for the period and solve for the frequency by taking the reciprocal of the period. This would give us the frequency of the small oscillations of the rod and springs system.

In summary, as a scientist, I would approach this problem by clearly defining the key variables and equations involved, determining the effective mass of the system, and then using the equation for the period of a simple harmonic oscillator to calculate the frequency of the oscillations.