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Equation for Simple Harmonic Motion?

  1. Jan 19, 2016 #1
    In Simple Harmonic Motion,

    (k/m) = ω2

    be expressed for all SHMs or only the ones in which the mass due to which the SHM is being executed is performing a circular motion?

    Since for example, in the case of spring, there is no circular motion involved, so omega should not be defined for spring.

    Or if it is a general equation is there a proof to justify?

    How I understand this is....

    (k/m) = ω2

    can be derived by comparing
    F = -kx ⇒ (d2x/dt2) = (-k/m)
    x = Acos(ωt + φo) ⇒ (d2x/dt2) = -ω2x

    But is
    x = Acos(ωt + φo)
    a general equation for all SHMs? That is to say, can all SHMs be expressed by this equation or only the ones in which circular motion is involved?

    For example,
    how would this equation be used for expressing the SHM performed by a spring?
    The above equation is using a sinusoidal function and is dealing with angles and there is no angle formed in SHM performed by spring since the SHM is being executed linearly. (The mass because of which SHM is being executed is moving linearly and also performing the SHM)

    Or if the equation is a general statement how can we theoretically prove it?
    Last edited: Jan 19, 2016
  2. jcsd
  3. Jan 19, 2016 #2
    The equation for S.H.M. ##\text{d}^2x/\text{d}t^2 + \omega^2 x =0## (where ##\omega^2=k/m##) does apply for a spring, as long as it doesn't experience damping or forcing.

    The difference is deriving the equation of motion. We use the fact that the resultant force on a body is the sum of all forces it experiences and assume the only force acting is the restorative force of the spring itself (see Hooke's law). We assume one end of the spring is anchored at the origin, the other end lies on the positive ##\textbf{i}##-axis [and that's the end we're considering] so end up with;
    $$\textbf{F}=m\textbf{a}=-kx\textbf{ i}$$
    where ##x## is the spring's displacement from its equilibrium position. This becomes the same scalar equation you use above.

    So S.H.M also describes the motion of an un-damped, un-forced model spring. The problem is when you add forcing or damping you get a more complicated equation of motion, so a different solution.
    For example, a forced damped harmonic oscillator has the equation
    $$m \frac{\text{d}^2x}{\text{d}t^2} + r \frac{\text{d}x}{\text{d}t} + kx = P \cos{(\Omega t)}$$
    where ##r## is the damping constant and ##P## & ##\Omega## are the amplitude & angular frequency of the forcing term.
  4. Jan 19, 2016 #3
    No actually I am only referring to an undamped unforced spring.

    The question is
    how can ω be defined for a spring when there is no circular motion involved? And so can we write
    (k/m) = ω2
    for all SHM's or only some specific ones?
    Like for SHM in spring what would ω refer to in lay man's language if we try to visualize it?
    similarly how can we visualize phase 'φ' for spring?
    In the case uniform circular motion, the foot of perpendicular from the body, on the x and y axes (assuming the origin to be the centre) performs SHM. Here we can visualize that ω as the angular speed and φ as the angle made by the body with the axis (say x axis). (let initial phase be 0 for instance, to avoid complications).
    Let the radius of the circular orbit be A.
    The angle made (phase) at any time t will be
    φ = ωt
    Resolving on x and y axes we get...
    x = Acos(ωt)
    y = Asin(ωt)

    on differentiating twice, we get the differential form of SHM equation...
    {of the form (d2x/dt2) + (kx/m) = 0}
    Which is a proof that it represents SHM. But not that all SHM equations can be expressed thus.

    But in the case of spring there is no circular motion, the body itself is performing SHM. So what would ω and φ refer to in that case? Since there is no angular motion and any angle formed...
  5. Jan 19, 2016 #4


    User Avatar

    Staff: Mentor

    The x- and y- projections (components) of uniform circular motion are simple harmonic motion:

  6. Jan 19, 2016 #5

    Yes that is what I referred to...
  7. Jan 19, 2016 #6


    User Avatar

    Staff: Mentor

    ##f = \frac{\omega}{2\pi}## is the frequency (rate of oscillation in cycles per second). If you consider 1 cycle as equivalent to ##2\pi## radians, than ##\omega## is the angular frequency (rate of oscillation in radians per second). Even though there is no physical angle involved in the motion of a spring, we nevertheless call it "angular frequency" because of the geometrical and mathematical relationship to uniform circular motion. Likewise with the variation of the electric and magnetic field in an electromagnetic wave; or the variation of the ##\Psi## function in quantum mechanics, which is even more abstract.
  8. Jan 19, 2016 #7
    Sorry, my mistake.

    ##\omega^2=k/m## does stand for all cases of SHM. For the spring, it's derived from the vector equation I gave which comes from Newton's laws and Hooke's law. So it stands for all cases where oscillations can be modeled as if a particle was moving in 1D solely under the influence of a model spring.

    There's not really a layman's way of interpreting ##\omega## for the case of a spring. As jtbell says ##f## is the cycles per second. The way I think of it is that we need ##\omega=2 \pi f## to make the maths work, since it's the argument of a sine function. That way, if ##f=1\text{ s}##, the function ##\sin{\omega t}## returns to its original value after 1 second.

    The phase ##\phi## moves the zero of the sine function left or right when looking at it's graph. This means the oscillations, described by ##x(t)=A\sin{\omega t +\phi}##, start (at ##t=0##) displaced from the origin.

    Like jtbell said, its the mathematical and geometrical relationship to uniform circular motion that's important - all the oscillatory motions mentioned are described by the family of sine functions.
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