Simple Pendulum SHM Problem: Calculating Angular Frequency for Maximum Amplitude

AI Thread Summary
The discussion focuses on calculating the angular frequency of a pendulum subjected to a variable horizontal force. The user initially struggled with the problem but applied the formula for the angular frequency of a simple pendulum, resulting in a correct answer of 5.6 rad/s. The response clarifies that this method is valid because the frequency used is the natural frequency of the system. It explains that systems oscillate with maximum amplitude when driven at their natural frequency, a phenomenon known as resonance. Understanding this principle helps clarify why the initial approach yielded the correct result despite the external force.
Hernaner28
Messages
261
Reaction score
0

Homework Statement


attachment.php?attachmentid=48776&stc=1&d=1341063270.gif

Consider, as shown in the picture, a penulum of mass m=0.52g, hanging on an ideal rope of length l=0.31m. A force is exerted to this pendulum as shown in the picture.
The horizontal component of the force is:
\displaystyle {{F}_{x}}={{F}_{0}}\cos \left( \omega t \right)
\displaystyle {{F}_{0}}=2.3N

Calculate the angular frequency that will make the system oscilate with the maximum amplitud. Consider small oscilations around the equilibrium position.

Homework Equations





The Attempt at a Solution



I did nothing beucase I don't know how to deal with that variable force. BUT, what I did was to use the angular frequency of a simple pendulim which is
\displaystyle \sqrt{\frac{g}{L}}
And I just replaced the values and I got the correct answer! 5.6rad/s.
But my question is: why did I get the correct result if there's a force there?

Thanks!
 

Attachments

  • spend.gif
    spend.gif
    5.9 KB · Views: 622
Physics news on Phys.org
Hi Hernaner28! :smile:

Your method works because the frequency you used is the natural frequency of the system. And anybody vibrates with a greater amplitude when externally forced to vibrate with its natural frequency(Resonance)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top