Instantaneous response of damped simple harmonic motion

AI Thread Summary
The discussion focuses on modeling simple harmonic motion (SHM) in Simulink, specifically addressing the delayed response of the system to input force changes. The user seeks to achieve a response time of less than 0.5 seconds, currently limited to 1.5 seconds due to damping and spring constant values. Key insights include the relationship between frequency, mass, and spring constant, with the formula w = sqrt(k/m) being essential for calculating the required spring constant for a desired frequency. Suggestions include reducing damping and adjusting spring constants, with a proposed value of approximately 400 to achieve the desired frequency. The conversation highlights potential issues in the simulation setup if increasing the spring constant leads to instability.
james6008
Messages
14
Reaction score
0
Hi

I am trying to model SHM in Simulink as shown here:
http://pundit.pratt.duke.edu/wiki/Simulink/Tutorials/DiffEq

I have tried using different values of spring constant and damping to get instant response to the input force. I am measuring the displacement calculated by SHM. The force changes with time and the model responds to the change but the response is delayed by a certain amount of time which depends on damping mostly but sometimes its due to spring constant too. I can not get the model to respond any quicker than 1.5 seconds. I would like it to respond in < 0.5s. I have mass of 5kg, force about 30N/m, spring constant of 35 and damping of 15. I am allowed to change the spring constant and damping as I like.

Any idea what could be causing the problem?
 
Physics news on Phys.org
Do you know how the frequency of the response in SHM depends on the mass and stiffness?

If you don't know the "formula", try setting the damping to a small value like 1 and see what happens with different spring constants.

If you do know the formula, work out what spring constant you need to get the frequency high enough. If you want a response in < 0.5 sec the frequency needs to be > 2 Hz.
 
Hi

I believe the formula you are talking about is:

w = sqrt (k / m )

m = 5kg and I can re-arrange this to calculate the spring constant (k) for 2 Hz. I get a value of 790 which is way too high. In order to get a response of 1.5s I am using k=45.

Another problem with the model is that if you increase the spring constant too much, the system never stabilises and the response starts going in the opposite direction infinitely.
 
In your OP you said you could have any spring constant you like. Now you changed the rules and said 790 is too high!

If you are getting the response you want in 1.5 sec, but you want to speed it up to 0.5 sec, that is multiplying the frequency by 3. So using the formula you need to multiply the stiffness by 32 = 9 which would give k = about 400.

If your response goes to infinity when k is high, there must be something else wrong with your simulation.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top