Frequency of oscillation of a rod with two springs

In summary, the conversation discusses the equilibrium state of two elongated springs and a rod. The springs are assumed to be elongated along the x-axis and the rod makes an angle with the vertical axis. The equation of motion for the system is derived and two different solutions are presented, with the second one being deemed correct after considering torque and force analysis.
  • #1
Pushoam
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Homework Statement



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Homework Equations

The Attempt at a Solution


Let's take that in equilibrium, both springs S_l and S_r are elongated by ##x_e##.
Assuming that the oscillations are so small that the springs could be taken along x- axis approximately.
At time t, S_l is elongated by x. The rod makes an angle ## \theta ## with vertical axis such that
## \tan \theta = \frac x {\frac l 2 }
##
This gives that the S_r will be elongated by x' = ## 2 \tan \theta = 2x ##

Eqn. of motion of com gives,
## k ( x_e + x ) (-\hat x ) +k (x_e -2x) \hat x = m \ddot x \hat x ##
## -3kx = m\ddot x##
## \omega = \sqrt (\frac {3k} m )##

Is this correct so far?
 
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  • #2
Another attempt:
Motivated from the OP's solution,
Torque eqn. about the pivot-point,
## k ( x_e + x ) \frac l 2 (-\hat z ) +k (x_e -2x) l \hat z= I \ddot \theta \hat z ##
But in equilibrium, the above eqn. gives me non-zero net torque, so it means that the elongation is not same for both the springs.
Let's say that in equilibrium, LHS elongation is ##x_{le} ## and RHS elongtion is ## x_{re}##, and ## x_{le } ≠ x_{re} ## . But these elongations are such that the net torque on the system about the pivot is zero in the equilibrium
But, this tells that net force acting on the com is non - zero in equilb. { A CONTRADICTION ?}

## \tan \theta = \frac {2x } l
\\ \text{ for small , } \theta ,
\\ \tan \theta≈ \theta
\ddot \theta = \frac {2 \ddot x} l##
So,
## k x \frac l 2 (-\hat z ) +k ( -2x) l \hat z= I \ddot \theta \hat z = \frac {ml^2}3 (\frac {2 \ddot x} l) ##
## \frac { -5} 2 kx = \frac 2 3 m \ddot x ##
## \omega = \sqrt {(\frac {15 k} {4 m}) }##

But how to decide the correctness of both the solutions?
 
  • #3
Pushoam said:
in equilibrium, both springs S_l and S_r are elongated by xe.
Is that possible? What would the two torques about the pivot be?
 
  • #4
haruspex said:
Is that possible?
No, it is not.
In the OP, I thought that it were only springs which were exerting forces along x-axis. Consequently, forces by them should be equal in magnitude in eqbm and so their elongation.
But, torque analysis in post #2 says that it is not so.
SO, I think the HINGE also exert force along x-axis to make the center of mass in eqbm. I missed it.

LEARNING: IN CASE OF RIGID BODY MOTION, I SHOULD DO BOTH TORQUE AND FORCE ANALYSIS BEFORE GOING AHEAD.
So, the solution in post #2 is correct.
Is this correct so far?
 
  • #5
Pushoam said:
the solution in post #2 is correct.
Yes.
 
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  • #6
Thanks.
 

FAQ: Frequency of oscillation of a rod with two springs

What is the frequency of oscillation of a rod with two springs?

The frequency of oscillation of a rod with two springs is determined by the mass of the rod, the spring constants of the two springs, and the length of the rod. It can be calculated using the formula: f = 1/2π√(k/m), where f is the frequency, k is the combined spring constant, and m is the mass of the rod.

How does the mass of the rod affect the frequency of oscillation?

The frequency of oscillation is directly proportional to the square root of the mass of the rod. This means that as the mass of the rod increases, the frequency of oscillation decreases. In other words, a heavier rod will have a lower frequency of oscillation compared to a lighter rod with the same spring constants.

What happens to the frequency if one of the springs is removed?

If one of the springs is removed, the frequency of oscillation will change. The new frequency can be calculated by using the formula: f = 1/2π√(k/m), where k is the remaining spring constant and m is the mass of the rod. The frequency will decrease if the spring constant decreases, and increase if the spring constant increases.

Can the frequency of oscillation be changed by adjusting the length of the rod?

Yes, the frequency of oscillation can be changed by adjusting the length of the rod. The frequency is inversely proportional to the square root of the length, meaning that as the length of the rod increases, the frequency decreases. This can be seen in the formula: f = 1/2π√(k/m), where m is the mass of the rod and k is the combined spring constant.

How does the spring constant affect the frequency of oscillation?

The spring constant has a direct effect on the frequency of oscillation. As the spring constant increases, the frequency increases. This can be seen in the formula: f = 1/2π√(k/m), where k is the combined spring constant. A higher spring constant means that the springs are stiffer, which results in a higher frequency of oscillation.

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