Unexpected result with bar fixed to spring with periodic loading

In summary, the conversation discussed the solution to a steady state problem involving a bar that is fixed to a string on one side and periodically pulled on the other side. To check the solution, the speaker made the young's modulus go to infinity, but the expression they received had an unexpected term that depends on frequency. They checked the units and calculations, but are unsure of the effect of frequency on the displacement field for a rigid bar. They also mentioned the model they used for the bar and the possibility of a rotary inertia term. They were advised to look up the theory of single-degree-of-freedom vibrating systems.
  • #1
ENgez
75
0
Hi, I solved a steady state problem involving a bar fixed to string in the left side and pulled periodically on the right side [itex]f(x,t)=P_0sin(wt)[/itex]. To check the solution i made E (young's modulus) go to infinity, essentially making the bar rigid. the expression i expected to receive is:

u(x,t) = [itex]\frac{P_0sin(wt)}{k}[/itex]

which is hookes law.

but the expression i received was:

u(x,t) = [itex]\frac{P_0sin(wt)}{k-ρ_{1D}Lw^{2}}[/itex]

the density is one dimensional and L is the bar length.
this expression has an extra term that depends on the frequency which subtracts from the spring constant.
i checked the units and my calculations and they seem to add up. i can't visualize the effect of frequency on the displacement field for a rigid bar. Does this term really "exist" or is this some kind of error?

BTW, the model i used for the bar is the longitudinal displacement equation for bars:
[itex](AEu)''+f(x,t)=ρ\stackrel{..}{u}[/itex]
 
Engineering news on Phys.org
  • #2
The "extra" term looks like the rotary inertia of the beam, and a term something like that should be there.

Look up the theory of single-degree-of-freedom (SDOF) vibrating systems.
 
  • #3
Thank you, I see it know.
 

1. What is the cause of the unexpected result?

The unexpected result could be caused by a number of factors, such as improper calibration of the equipment, external forces acting on the system, or errors in data collection or analysis.

2. How can the unexpected result be explained?

The unexpected result can be explained through further experimentation and analysis. It is important to consider all possible variables and to repeat the experiment multiple times to ensure accuracy.

3. Can the unexpected result be replicated?

It is possible to replicate the experiment to see if the same unexpected result occurs. However, it is important to take note of any changes in equipment or environmental conditions to determine if they may have an impact on the results.

4. How can the unexpected result be prevented in future experiments?

To prevent unexpected results in future experiments, it is important to carefully plan and design the experiment, calibrate equipment properly, and control for external factors. It may also be helpful to consult with other experts in the field for advice and guidance.

5. What are the implications of the unexpected result?

The unexpected result may have implications on the validity of the experiment and the overall understanding of the phenomenon being studied. It is important to carefully analyze the unexpected result and determine if it requires further investigation or if it can be explained and accounted for in future experiments.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
203
Replies
23
Views
1K
  • Classical Physics
Replies
0
Views
146
  • Introductory Physics Homework Help
Replies
24
Views
1K
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
377
  • Classical Physics
Replies
17
Views
1K
Replies
1
Views
517
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top