Calculating Displacement of Uniform Bar at Centre with Respect to Time

In summary, a uniform bar of length "l" and mass "m" is supported at its ends by two springs of different stiffness. When pulled and released from its equilibrium position, the centre of gravity of the bar will oscillate with a complicated motion, which can be solved using normal modes and coupled differential equations. This problem becomes even more complex when considering different scenarios, such as equal or unequal springs, parallel and series arrangements, and varying amplitudes of bar motion. Asking the right questions is crucial in finding the correct solution.
  • #1
Numbers123
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A uniform bar of length "l" and mass "m" is supported at its ends by two springs. One spring has twice the stiffness of the other. From the equilibrium position the centre of the bar is pulled down a short distance and then released. How do you calculate the displacement "s" of the centre of gravity with respect to time?
This is a self inflicted question after looking at the vibrations of weights on springs. I am not a student (63 years old) just interested in physics.
I am assuming that the springs at either end of the bar will oscillate with simple harmonic motion of different frequencies (one twice the other) but how do they combine at the centre.
 
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  • #2
That problem is a bit complicated. It as to be solved using "normal modes", which follow from coupled differential equations.
 
  • #3
Numbers123 said:
A uniform bar of length "l" and mass "m" is supported at its ends by two springs. One spring has twice the stiffness of the other. From the equilibrium position the centre of the bar is pulled down a short distance and then released. How do you calculate the displacement "s" of the centre of gravity with respect to time?
This is a self inflicted question after looking at the vibrations of weights on springs. I am not a student (63 years old) just interested in physics.
I am assuming that the springs at either end of the bar will oscillate with simple harmonic motion of different frequencies (one twice the other) but how do they combine at the centre.

This is a very complicated problem you have self-inflicted. But it's interesting to think about.

First just think about some special cases. Consider the case where both springs are the same. What happens when springs are in parallel? What would be the equivalent spring constant? How is it different from springs in series? What would be the spring constant then? What happens if two springs are replaced by one spring at the center of the bar? How does the bar oscillate if it is pulled and released without tilting it? Can you use the concept of "center of mass" then to assume all the mass of the bar is at one point? What happens if you have two equal springs and you pull the bar equally and release it. What happens if you have two equal springs on the end and you lower one side and raise the other side and release it? Are the resonant frequencies the same in both cases? (springs and mass have not changed) Explain the result. What happens if the springs are not equal? Can the motion of each end of the bar be separated from that of the other end? Does it help if one assumes the vibrations of the bar are of very small amplitude. What happens to your forces as the amplitude of the bar motion gets very large (has to do with the angles of the forces a the point of spring attachments)?

Do you see what a can of worms this "simple" problem is? Getting the right answer is going to be all about asking the right questions.
 

FAQ: Calculating Displacement of Uniform Bar at Centre with Respect to Time

What is the formula for calculating displacement of a uniform bar at the center with respect to time?

The formula for calculating displacement of a uniform bar at the center is d = (1/2)*a*t^2, where d is displacement, a is acceleration, and t is time.

How do you determine the acceleration of a uniform bar?

The acceleration of a uniform bar can be determined by dividing the change in velocity by the change in time, or by using Newton's second law of motion, F=ma, where F is the net force acting on the bar and m is its mass.

Can displacement be negative when calculating the displacement of a uniform bar at the center?

Yes, displacement can be negative when calculating the displacement of a uniform bar at the center. This indicates that the bar is moving in the opposite direction of its initial position.

What are the units for displacement?

The units for displacement are typically in meters (m) or centimeters (cm). However, it can also be measured in other units such as feet (ft) or inches (in).

Do I need to know the initial velocity to calculate displacement of a uniform bar at the center?

Yes, the initial velocity is needed to calculate displacement of a uniform bar at the center. This is because displacement is a measure of an object's change in position, which includes both distance and direction. Without knowing the initial velocity, the direction of the displacement cannot be determined accurately.

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