A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 4 years ago

SHM question: Find time period:

  • This Question is Closed
  1. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1328424655814:dw|

  2. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The two pulleys are fixed...

  3. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The period of a simple set-up is expressed as\[T = 2 \pi \sqrt{m \over k}\]Assuming the mass is constrained to vertical motion, the period is expressed as\[T = 2 \pi \sqrt{m \over 2k}\]

  4. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How do you say that? If you move the block down by x. The springs dont extend by the same x. So F is not equal to 2kx.

  5. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The period doesn't depend on the displacement.

  6. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if extension in springs is not x then force is not 2kx. which clearly implies the constant isn't 2k. If the constant isn't 2k how do you say that period remains same.

  7. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well if u see in the formula of time period there is not saying abu distance i guess

  8. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    T=2pie under root m/k

  9. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    means it is independent from distance

  10. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    But there is k which is different.

  11. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i agree it will be different i m just telling in response to that distance u were saying

  12. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Heena is right. Period of oscillation is isochorous. The spring constant doesn't change either.

  13. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea and acc. to the ques i saw they didnt said K1 or K2 means it can be same also if there is no stratement

  14. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    heena cm there

  15. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well if you still insist your answer is not there in the options.

  16. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    See the spring constant may be same but it depends on the force The force is NOT 2kx.

  17. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is that figure complete? We can analyze the kinematics but we'll need more information. The angle theta changes with displacement.

  18. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thats all is given. i think slant can be kept constant.

  19. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is this an option?\[T = 2\pi \sqrt{2 \over 2 k \cos(\theta)}\]

  20. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Rather. \[T = 2 \pi \sqrt{m \over 2k \cos(\theta)}\]

  21. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cos theta is outside the sqrt.

  22. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I would go with that one. Sorry for the bad answer.

  23. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Could you explain?

  24. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    dont u think shankvee its a mixing question of NLM and SHM i mean here spring is going down means pulling is rolling down so tension is also there as COME i think we should compare them

  25. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Here tension=force of spring. There is no mixing of NLM and all... I just don;t get if you pull the mass by x how much the springs get extended. Eashmore: Is the extension in the springs x cos theta or is it x/ cos theta , Could you explain this, i never get this.

  26. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    as spring is extended by x it got comression too x but here if u see in figure there is string too

  27. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No it is not x. If you pull in one direction by x then the extension in another direction can't be x how is it possible?

  28. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nah i m sayin if u extend by x then the compression will bw of x luk in diagram|dw:1329036256383:dw| A=expansion b-contraction

  29. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    True but the expansion isn't what i'm saying is.|dw:1328429223949:dw| If x is amount pulled downward x is not equal to y.

  30. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    t=F right look at this figure that i m hoping abu this ques to be |dw:1329037050794:dw|

  31. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2Tcos thta-mg=ma right?

  32. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now if we arrege it to form of shm equation T yeilds

  33. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how u take 2T i mean here box applies T' and pulley is apllying T

  34. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    even i m not sure how to solve i only know how to make diagrams and wat can be done here but how to calculate for that case i say help :P

  35. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is a out come of vectors also T=T'

  36. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    T in newton formula is tntion & next is period of time

  37. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    don't get my goal yea?

  38. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok thenn so by this we can say na that angle between them is120 degree ?? if manitude same and resulatant is also same then the angle must be 120degree

  39. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    na?

  40. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what is mean?

  41. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lolz its a word of hindi just tongue slip forget it :P

  42. Mani_Jha
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If we take the spring's extension be x, then the block moves by \[x \cos \theta\]

  43. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *bookmark. I have to go now, but if this is still unanswered when I'm back I'll help.

  44. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok thanks a lot...

  45. Mani_Jha
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    For the block of mass m, \[2T \cos \theta=-ma\] T=kx So, \[2k \cos thetax+ma=0\] This is a differential equation in x, a=d2x/dt2 So, its solution will be \[T=2\pi \sqrt{m/2k \cos \theta}\] You have to just take the coefficients of a and x. That should be the answer as far as i think

  46. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.