1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple Harmonic Motion

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Natural length is "l" and when a "m" mass is hanged on the P edge, it dragged by another "L" length. Point "O" is "4L" far from the floor.
    1. Find [itex]\lambda[/itex]
    2. The mass will be hold at point "O" and released with the starting velocity √(gL). Find the velocity of the mass after it reach "L" distance.
    3. Prove that f + [itex]\frac{g}{L}[/itex]x = 0. ( where "f" = acceleration)
    4. From the above equation, assume that v2 = [itex]\frac{g}{L}[/itex][ c2 - x2 ] and find the value of "c". (V = velocity)
    5. Show that the velocity of mass "m" is zero when it comes to the ground and show that the total time to reach the floor from the point "O" is equal to [itex]\frac{1}{3}[/itex] (3[itex]\sqrt{3}[/itex] - 3 + 2∏)[itex]\sqrt{\frac{L}{g}}[/itex]

    2. Relevant equations
    1. Tension = Weight [T = mg]
    2. m[itex]\frac{dv}{dt}[/itex] = mg - T
    3. T = [itex]\lambda[/itex][itex]\frac{e}{L}[/itex]
    4. ω = 2∏f ( where "f" = frequency )

    3. The attempt at a solution
    I can do half of the problem. And these are the answers.
    i . [itex]\lambda[/itex] is mg
    ii. Velocity after reaching "L" length = [itex]\sqrt{3gL}[/itex]
    iii. Can prove that f + [itex]\frac{g}{L}[/itex]x = 0. ( where "f" = acceleration)
    iv. Can prove that velocity is zero when the mass reaches the floor.

    I can't find a proper value for the constant "c" and prove the time equation.
     

    Attached Files:

    • 1.JPG
      1.JPG
      File size:
      3.4 KB
      Views:
      62
  2. jcsd
  3. Jun 3, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Here's some help for finding the constant c. For reference, let point B be the point at distance L below point O. Between O and B the mass is in free fall. So, you should be able to find the speed of the mass at B as a free-fall problem. After that, the mass is in SHM and the equation with c in it will apply. Think about the value of x at point B. Then apply the formula with c to the point B.
     
  4. Jun 3, 2012 #3

    TSny

    User Avatar
    Homework Helper
    Gold Member

    [I see that you already found the speed at B. Sorry.]
     
  5. Jun 3, 2012 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Prove that the first two terms in the answer for the time correspond to the time between O and B. From B to the floor the motion is SHM. Try using the equation of motion for SHM (x as a sinusoidal function of time) to find the time when the mass is at B and the time when the mass is at the floor. For the SHM, I would suggest letting t = 0 correspond to the mass passing through x = 0 (the equilibrium point of the SHM).
     
  6. Jun 3, 2012 #5
    Thank you TSny for you reply
    I tried to find the constant "c" by applying the following values.
    i. When x = 0 ; v = [itex]\sqrt{3gL}[/itex]
    So I'm getting a value of "[itex]\sqrt{3}[/itex]L"​
    as "c"

    The problem comes when drawing the correspondent circular motion to the SHM. I can see the amplitude is [itex]\sqrt{3}[/itex]L. Can you please draw it for me??? I've drawn what I can think of. Is that "2L" distance from point "o" the equilibrium??? and can I consider it as the center (as C in the picture) of the circle???
     

    Attached Files:

    • 2.JPG
      2.JPG
      File size:
      7 KB
      Views:
      42
    Last edited: Jun 3, 2012
  7. Jun 3, 2012 #6
    And if I apply "2L" from the point "o" as the equilibrium, I'm getting cos ∂ = √3
    That value is making it hard to get the final answer of that time function.
     
  8. Jun 4, 2012 #7

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Note that the equilibrium position of the SHM is the point where the mass would hang at rest; namely, at distance 2L below O. Thus, x = 0 corresponds to 2L below O. If you measure x positive downward, then the point located a distance L below O would correspond to x = -L. At distance L below O you have found the velocity to be √(3gL). So, try using this information to find the correct value for c.

    Then, you'll need to show that the mass comes to rest at the floor. What does this fact imply for the value of the amplitude of the SHM? What is the value of x at the floor?

    Finally, you'll need to use your SHM reference circle to find the time to go from x = -L to the value of x at the floor.
     
  9. Jun 6, 2012 #8
    Thanks for the info. I was wrong at calculating "C" then :D I can manage after that. Thanks again!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple Harmonic Motion
Loading...