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Under-damped simple harmonic motion solution derivation

  1. Aug 30, 2014 #1
    I have learnt in 1st year that the under-damped simple harmonic motion can be described by the differential equation [tex] m \frac {d^2 x} {dt^2} + b \frac {dx} {dt} + kx = 0 [/tex] where [itex] m [/itex] is the mass, [itex] b [/itex] is the constant of linear drag and [itex] k [/itex] is the spring constant

    But the derivation is skipped and only the solution is given, which is [tex] x(t) = x_m e^{-\frac {bt} {2m}} \cos(\omega' t + \phi)[/tex] Recently, I have learnt Laplace transform on YT, so I try to come up with my own derivation, which is as followed.
    [tex] m \frac {d^2 x} {dt^2} + b \frac {dx} {dt} + kx = 0 [/tex]
    [tex] m s^2 X(s) - m s x(0) - m x'(0) + b s X(s) - b x(0) + k X(s) = 0 [/tex]
    Let the initial displacement [itex] = x(0) = x_0 [/itex], initial velocity [itex] = x'(0) = 0 [/itex]
    [tex] m s^2 X(s) - m s x_0 + b s X(s) - b x_0 + k X(s) = 0 [/tex]
    [tex] X(s) (m s^2 + b s + k) = x_0 (m s + b) [/tex]
    [tex] X(s) = x_0 \frac {m s + b} {m s^2 + b s + k} [/tex]
    [tex] X(s) = x_0 \frac {s + \frac {b} {m}} {s^2 + \frac {b} {m} s + \frac {k} {m}} [/tex]
    Complete the square,
    [tex] X(s) = x_0 \frac {s + \frac {b} {m}} { (s^2 + \frac {b} {m} s + \frac {b^2} {4 m^2}) + \frac {k} {m} - \frac {b^2} {4 m^2}} [/tex]
    [tex] X(s) = x_0 \frac { (s + \frac {b} {2m}) + \frac {b} {2m} } { (s + \frac {b} {2m})^2 + \frac {k} {m} - \frac {b^2} {4 m^2} } [/tex]
    [tex] X(s) = x_0 \left [ \frac { s + \frac {b} {2m} } { (s + \frac {b} {2m})^2 + \frac {k} {m} - \frac {b^2} {4 m^2} } + \frac { \frac {b} {2m} } { (s + \frac {b} {2m})^2 + \frac {k} {m} - \frac {b^2} {4 m^2} } \right ] [/tex]
    Apply the First Shifting Theorem,
    [tex] x(t) = x_0 e^{- \frac {b} {2m} t } \cos \left ( \sqrt{\frac {k} {m} - \frac {b^2} {4 m^2}} t \right ) + x_0 \frac {b} {2m} \frac {1} { \sqrt { \frac {k} {m} - \frac {b^2} {4 m^2} } } e^{- \frac {b} {2m} t } \sin \left ( \sqrt{\frac {k} {m} - \frac {b^2} {4 m^2}} t \right ) [/tex]
    [tex] x(t) = x_0 e^{- \frac {b} {2m} t } \cos \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) + x_0 \frac {b} {2m} \frac {2 m} { \sqrt{4 k m - b^2} } e^{- \frac {b} {2m} t } \sin \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) [/tex]
    [tex] x(t) = x_0 e^{- \frac {b} {2m} t } \cos \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) + x_0 \frac {b} { \sqrt {4 k m - b^2} } e^{- \frac {b} {2m} t } \sin \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) [/tex]
    To convert the answer into the form [itex] x_m e^{-\frac {bt} {2m}} \cos(\omega' t + \phi) [/itex], we must compute the hypotenuse of the triangle with side [itex] 1 [/itex] and [itex] \frac {b} { \sqrt {4 k m - b^2} } [/itex]
    [tex] \sqrt{ 1^2 + \frac {b^2} {4 k m - b^2} } [/tex]
    [tex] = \sqrt{ \frac {4 k m - b^2 + b^2} {4 k m - b^2} } [/tex]
    [tex] = 2 \sqrt{ \frac {k m} {4 k m - b^2} } [/tex]
    [tex] x(t) = 2 x_0 e^{- \frac {b} {2m} t } \sqrt{ \frac {k m} {4 k m - b^2} } \left [ \frac {1} { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } \cos \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) + \frac { \frac {b} { \sqrt {4 k m - b^2} } } { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } \sin \left ( \frac { \sqrt{4 k m - b^2} } {2 m} t \right ) \right ] [/tex]
    If [itex] \cos \phi = \frac {1} { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } [/itex], then [itex] \sin \phi = \frac { \frac {b} { \sqrt {4 k m - b^2} } } { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } [/itex]

    By [itex] \cos(A) \cos(B) + \sin(A) \sin(B) = \cos(A-B) [/itex],
    [tex] x(t) = 2 x_0 e^{- \frac {b} {2m} t } \sqrt{ \frac {k m} {4 k m - b^2} } \cos \left [ \frac { \sqrt{4 k m - b^2} } {2 m} t - \arccos \left ( \frac {1} {2} \sqrt{ \frac {4 k m - b^2} {k m} } \right ) \right ] [/tex]
    Is the derivation correct?
    I am pretty confident with [itex] \omega' = \frac { \sqrt{4 k m - b^2} } {2 m} [/itex]
    However, I am not so sure about the amplitude and the phase. Is the following true?
    [tex] x_m = 2 x_0 \sqrt{ \frac {k m} {4 k m - b^2} } [/tex]
    and
    [tex] \phi = \arccos \left ( \frac {1} {2} \sqrt{ \frac {4 k m - b^2} {k m} } \right ) = \arcsin \left ( \frac { \frac {b} { \sqrt {4 k m - b^2} } } { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } \right ) = \arcsin \left ( \frac {b} {2 \sqrt{k m} } \right ) [/tex]
    Finally, is there an easier way to come up to a solution? Doing these seems to be very exhausting and prone to careless mistakes. Also, I want to learn about over-damping and critical damping. Isn't the same differential equation apply? What are their differences? Thanks for reading my question! I know it is long!
     
  2. jcsd
  3. Aug 30, 2014 #2

    Philip Wood

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    Sorry, haven't checked your working.

    But there is an easier way, and it's the standard method! Start by replacing the constants by different ones that will be easier to work with. Specifically, put [itex]\frac{b}{m}=2c[/itex] and put
    [itex]\frac{k}{m}=\omega_0^2[/itex]. [You should recognise that [itex]\omega_0[/itex] is the angular frequency of the system if b = 0 (no damping).

    At this stage we need to know if you've worked with complex numbers.The neatest method (imo) uses these. It starts by noting that [itex]x = Ae^{\lambda t}[/itex] must fit the equation, because if we make this substitution, x comes out as a common factor of all three terms, so the equation will hold good for all values of t, provided that [itex]\lambda^2 + 2c\lambda + \omega_0^2 = 0[/itex]. I should have said that [itex]\lambda[/itex] is a constant. You can find it from the quadratic equation, which will have non-real roots if [itex]c < \omega_0[/itex]. At this stage you really do need to be able to manipulate complex numbers...
     
  4. Aug 30, 2014 #3

    AlephZero

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    I've seen that version in some textbooks, but I think its even better to let ##\frac k m = \omega_0^2## amd ##\frac b m = 2c\omega_0##. That has the advantage that ##c## is dimensionless (exercise for the OP, check that, using the dimensions of ##m##, ##b## and ##k##) and ##c < 1##, ##c = 1##, and ##c > 1## for under, critical, and over damping.

    In fact, you can go a step further, and let ##X(\omega_0 t) = x(t)##, which simplifies the equation from ##\ddot x + 2 c \omega_0 \dot x + \omega_0^2 x = 0## to ##\ddot X + 2 c \dot X + X = 0##. Physically, that just means you are measuring time in units of the angular frequency of the undamped vibration, not in seconds.
     
  5. Aug 31, 2014 #4

    Philip Wood

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    I rather like [itex]\frac{b}{m}=2c \omega_0[/itex], but using [itex]X(\omega_{0}t)[/itex] in place of x(t) is a sophistication too far for me. It's a matter of taste.
     
  6. Sep 6, 2014 #5
    Thanks a lot! It sounds silly but I haven't heard of the standard method yet. OK I will try to use substitutions followed by using complex number (Euler's equation is useful!) to get a neat and mistake-free derivation.
     
  7. Sep 6, 2014 #6

    Philip Wood

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    You can always come back and post again if you get stuck. Good luck!
     
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