# Under-damped simple harmonic motion solution derivation

1. Aug 30, 2014

### alexvong1995

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

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

By $\cos(A) \cos(B) + \sin(A) \sin(B) = \cos(A-B)$,
$$x(t) = 2 x_0 e^{- \frac {b} {2m} t } \sqrt{ \frac {k m} {4 k m - b^2} } \cos \left [ \frac { \sqrt{4 k m - b^2} } {2 m} t - \arccos \left ( \frac {1} {2} \sqrt{ \frac {4 k m - b^2} {k m} } \right ) \right ]$$
Is the derivation correct?
I am pretty confident with $\omega' = \frac { \sqrt{4 k m - b^2} } {2 m}$
However, I am not so sure about the amplitude and the phase. Is the following true?
$$x_m = 2 x_0 \sqrt{ \frac {k m} {4 k m - b^2} }$$
and
$$\phi = \arccos \left ( \frac {1} {2} \sqrt{ \frac {4 k m - b^2} {k m} } \right ) = \arcsin \left ( \frac { \frac {b} { \sqrt {4 k m - b^2} } } { 2 \sqrt{ \frac {k m} {4 k m - b^2} } } \right ) = \arcsin \left ( \frac {b} {2 \sqrt{k m} } \right )$$
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. Aug 30, 2014

### Philip Wood

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 $\frac{b}{m}=2c$ and put
$\frac{k}{m}=\omega_0^2$. [You should recognise that $\omega_0$ 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 $x = Ae^{\lambda t}$ 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 $\lambda^2 + 2c\lambda + \omega_0^2 = 0$. I should have said that $\lambda$ is a constant. You can find it from the quadratic equation, which will have non-real roots if $c < \omega_0$. At this stage you really do need to be able to manipulate complex numbers...

3. Aug 30, 2014

### AlephZero

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.

4. Aug 31, 2014

### Philip Wood

I rather like $\frac{b}{m}=2c \omega_0$, but using $X(\omega_{0}t)$ in place of x(t) is a sophistication too far for me. It's a matter of taste.

5. Sep 6, 2014

### alexvong1995

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.

6. Sep 6, 2014

### Philip Wood

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