A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Simple harmonic motion question
anonymous
 one year ago
Simple harmonic motion question

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it said that the particles acceleration is proportional to displacement but sign is opposite? What does that mean? I know that the displacement graph looks like this and the acceleration graph looks like this for SHM ....dw:1441423084425:dw The textbook also said that acceleration magnitude is zero when displacement is zero (which for velocity is the max) and acceleration is max when particle at the crest or trough (for velocity is 0). Is that what it means by sign is opposite?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix @UnkleRhaukus

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Yes, it means opposite sign :) In this example, I was trying to show that if you're displaced an amount \(\large\rm x\) from equilibrium, then the acceleration is \(\large\rm x\) at the same time. I'm not so good with the physics, it would be a lot easier to explain this using calculus lol oh well!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2the key to shm is the restorative force. All proper shm starts with the equation F = kx the restorative force F is proportional to displacement x but acts in the other direction [hence  sign], ie back toward the origin/point of equilibrium everything everything everything follows from that.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the harmonic motion can be modeled by this function: \[\Large x\left( t \right) = A\cos \left( {\omega t + \phi } \right)\] we can set \( \Large \phi=0 \), without loss of generality, so we can write: \[\Large x\left( t \right) = A\cos \left( {\omega t} \right) \qquad \qquad (1)\] of course, the constant \( \large A \) is the amplitude of our motion. Now taking the second derivative with respect to time of the last equation, we get: \[\Large \ddot x\left( t \right) =  A{\omega ^2}\cos \left( {\omega t} \right) =  {\omega ^2}x\left( t \right) \qquad (2)\] As you can see, your graph are the graphs of the equations \( (1), \; (2) \) respectively @dareintheren

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2neglecting the scale factor \( \Large \omega^2 \), of course

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2and from \[\Large \ddot x\left( t \right) =  A{\omega ^2}\cos \left( {\omega t} \right) =  {\omega ^2}x\left( t \right) \qquad (2)\] note that \[\Large m\ddot x\left( t \right) = F =  m{\omega ^2}x\] the **restorative** force

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2and just to be totally clear, the maths follows the idea of a restorative force, and not the other way round:p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you guys for responding, more clear now, I can only medal the best one, but all was good!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2see this too http://openstudy.com/study#/updates/55daa36de4b02663346bfab3
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.