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!

Transverse Oscillation

  1. Mar 6, 2007 #1
    A mass m is connected to two springs with equal spring constants k. In the horizontal position shown, each spring is streched by an amount [itex]\Delta a[/itex]. The mass is raised vertically and begins to oscillate up and down. Assuming that the displacement is small, and ignoring gravity, show that the motion is simple harmonic and find the period.

    Here is my analysis: The mass is streched vertically such that the springs strech an amount [itex]a + \Delta a[/itex]. The restoring force, F, is acting in the direction of the springs. The horizontal component of F from both springs cancel out but the vertical component from both springs are both in the same direction. Let [itex]F \sin \theta[/itex] be the vertical component of F, [itex]\theta[/itex] being the angle the string makes with the horizontal. The sum of the vertical forces acting on m, ignoring gravity is [itex]2 F \sin \theta[/itex].

    What is [itex]\sin \theta[/itex]? Some quick trig. yields:

    [tex]\sin \theta = \sqrt{(a + \Delta a)^2 - a^2} / (a + \Delta a) = \sqrt{2a \Delta a + (\Delta a)^2} / (a + \Delta a)[/tex]

    Since the displacement is small, it is "safe" to approximate [itex]\sin \theta[/itex] as:

    [tex]\sin \theta = \sqrt{2a \Delta a} / a = \sqrt{2 \Delta a / a} [/tex]

    And so the sum of the vertical forces on m is:

    [tex]2 F \sin \theta = 2 (-k \Delta a) \sqrt{2 \Delta a / a} [/tex]

    The above is not akin to Hooke's law so I'm wondering how in the world is the motion simple harmonic? Perhaps I did something wrong in my analysis...
     
    Last edited: Mar 6, 2007
  2. jcsd
  3. Mar 6, 2007 #2
    A picture is worth a 1000 words; I'm assuming that you have:
    spring--mass--spring under tension and horizontal, than displace the mass in the y direction, and it bobs up and down?
     
  4. Mar 6, 2007 #3
    if so, then the expression for force along the Y axis, Fy, is then:

    -2k(sin(theta))*l where l=length of string/spring. This could be also written as
    m*y"+2*k*y(t)=0.

    This is a second order ODE that describes periodic motion.

    Just like a bass guitar string. Intuitively I know that the original delta a will affect the tuning frequency, need a minute to figure out how mathematically..
     
    Last edited: Mar 6, 2007
  5. Mar 6, 2007 #4
    I believe the restoring force is 2sin(theta)*-2K*deltaA using the prestretched force figure and assuming small displacements about the equilibrium. In other words the result is the same as ypu have, this is still of the form,
    my"+Ky=0 where (kY)>1, and so described undamped harmonic motion.
     
  6. Mar 6, 2007 #5
    Yes, exactly. It's a shame that I don't have a scanner to reproduce the picture.

    As I understand it, if the restoring force obeys Hooke's law, i.e. if the restoring force is proportional to the product of the displacement and some constant and the force points in the direction of the equilibrium point, then the motion is simple harmonic. The problem as I see it is that the constant is not a constant (it depends on the displacement).
     
  7. Mar 6, 2007 #6
    But look at hooke's law which if you were to simply have one spring attached and put the mass horizontally on an airtable and pull against the spring, it would be periodic as in f=-kx.

    The eqn for position comes as a result of having conjugate imaginary roots e^ix and e-ix (the sqrty of x actually) which thru Eulers identity iirc, can be represented as sum of sin and cos, or lumped together in a single trig fx.
     
    Last edited: Mar 6, 2007
  8. Mar 6, 2007 #7
    That is my point. What is the force constant in

    [tex]2 F \sin \theta = 2 (-k \Delta a) \sqrt{2 \Delta a / a} [/tex] ?

    It is [itex] -2k \sqrt{2 \Delta a / a}[/itex] which is dependent on the displacement, i.e. it is not a constant.
     
    Last edited: Mar 6, 2007
  9. Mar 6, 2007 #8
    I see what you are asking now--the original displacement delta a. Sorry. If i understand the problem correctly, thats why the additional displacement is assumed small. So delta A is a "constant" in this case.
     
  10. Mar 6, 2007 #9
    That doesn't make sense. Just because the displacement is small, doesn't mean it is constant. If that were true, then the restoring force would be constant and there would be no SHM.
     
  11. Mar 6, 2007 #10
    No, not what I meant: the original stretch in line of springs, delta A, (the prestretch) is a constant, the transverse amplitude is both small and variable. Hope that helps.
     
  12. Mar 7, 2007 #11
    At an instance in time, any force is constant. If the amplitude is variable, i.e. it varies over time, then the motion is not simple harmonic. It may be damped harmonic motion or something else.

    My point is that the equation I get does not look like [itex]F = -kx[/itex]. It looks like [itex]F = -k(x) x[/itex] where k(x) is a function of x, the displacement, or it looks like [itex]F = -k x^{3/2}[/itex]. Hence, it is NOT simple harmonic.
     
  13. Mar 7, 2007 #12
    I see it more like a pendulum with tension 2*T where T=-k*Delta(a), and the restoring force= sin (theta) * T =x/L * T
     
  14. Mar 7, 2007 #13
    By the way, the period according the book's answer is [itex]2 \pi (ma/2k \Delta a)^{1/2}[/itex], but according to my calculations (see post 7), it would be

    [tex]2 \pi \sqrt{\frac{m}{ 2k \sqrt{2 \Delta a / a}}[/tex]

    Because of the difference, I'm wondering if my analysis is wrong to begin with.
     
  15. Mar 7, 2007 #14
    what if we were to multiply both top and bottom of the quantity within the radical by a.
     
  16. Mar 8, 2007 #15

    Doc Al

    User Avatar

    Staff: Mentor

    Here's how I would analyze this:

    Since you are trying to express the motion in terms of the vertical displacement, I will assume that [itex]\Delta a[/itex] is the vertical displacement of the mass. When the mass is displaced that amount, the springs are stretched to a distance: [itex]\sqrt{a^2 + (\Delta a)^2}[/itex].

    So the tension in each spring is now:

    [tex]T = k \sqrt{a^2 + (\Delta a)^2}[/tex]

    Thus the restoring force (the sum of the vertical components of tension from each spring) is now:

    [tex]F = 2 k \sqrt{a^2 + (\Delta a)^2}\frac{\Delta a}{\sqrt{a^2 + (\Delta a)^2}}[/tex]

    Which is just:

    [tex]F = 2 k \Delta a[/tex]

    The motion is clearly simple harmonic. (And it should make intuitive sense that the spring constant doubled, since you have two springs in parallel.)

    I don't understand this answer, since it implies that the period depends on the amplitude--that's not SHM. Seems like an error to me.
     
    Last edited: Mar 8, 2007
  17. Mar 8, 2007 #16
    I just dawned on me this morning. The displacement [itex]\Delta a[/itex] in my analysis is not the vertical displacement. The vertical displacement is [itex]\sqrt{(a + \Delta a)^2 - a^2}[/itex]. Let y be this vertical displacement. Then,

    [tex]\sin \theta = \frac{y}{a + \Delta a}[/tex]

    and so

    [tex]2F \sin \theta = 2 (-k \Delta a) \frac{y}{a + \Delta a} = - \frac{2ky \Delta a}{a + \Delta a} [/tex].

    For small [itex]\Delta a[/itex], the above is approximated by [itex]- (2k \Delta a / a)y[/itex]. This is akin to Hooke's law and so the motion is SHM. The period is

    [tex] 2 \pi \sqrt{\frac{m}{2k \Delta a / a}} = 2 \pi \sqrt{\frac{ma}{2k \Delta a}}[/tex]

    which does look like the books answer.
     
    Last edited: Mar 8, 2007
  18. Mar 8, 2007 #17
    nice job there.
     
  19. Mar 8, 2007 #18

    Doc Al

    User Avatar

    Staff: Mentor

    This still doesn't seem quite right to me. You are using [itex]a[/itex] to represent the amount that the springs are stretched when they are in there equilibrium position, and [itex]a + \Delta a[/itex] is the amount that the springs are stretched when the mass is displaced vertically. (Note: The mass moves vertically. If the mass exhibits SHM, the restoring force must be proportional to its vertical displacement from equilibrium.)
    Looks good.

    I would say that the spring tension is given by [itex]k (a + \Delta a)[/itex], not [itex]k \Delta a[/itex], so that:

    [tex]2F \sin \theta = -2 k (a + \Delta a) \frac{y}{a + \Delta a} = - 2 k y[/tex]

    That is SHM about the equilibrium position.

    Since [itex]\Delta a[/itex] and [itex]y[/itex] are not independent variables, this does not look like Hooke's law or SHM to me.

    One of the characteristics of SHM is that the period is independent of amplitude, which is not the case with the book answer. So I don't see how this answer can be considered SHM. (If I'm missing something obvious, let me know! :cool: )

    What text are you using?
     
  20. Mar 8, 2007 #19
    ignore, this/
     
    Last edited: Mar 8, 2007
  21. Mar 8, 2007 #20
    You're right about that. Hooke's law requires a force constant and since [itex]\Delta a[/itex] is not constant, it does not satisfy Hooke's law.

    I'm using "Physics for Scientists and Engineers with Modern Physics" by Douglas C. Giancoli: http://cwx.prenhall.com/bookbind/pubbooks/giancoli3/
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?