By Jimmy Apablaza
This problem is described in Exercise 14, Section 7.6 (page 323) of the A first Course in Differential Equations textbook, 8ED (ISBN 0-534-41878-3).
Figure 1. Coupled Spring System.
Problem Statement
Derive the system of differential equations describing the straight-line vertical motion of the coupled spring shown in Figure 1. Use Laplace transform to solve the system when
,
, and
,
,
, and
.
Solution
At positions
and
, the masses
and
are in equilibrium. Thus, the motion equations for
and
are,
![{\displaystyle m_{1}{\ddot {x}}_{1}=-k_{1}x_{1}+k_{2}(x_{2}-x_{1})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff2799f6339389d00beca01b77481ca8f3734960)
- ∴
![{\displaystyle m_{1}{\ddot {x}}_{1}+k_{1}x_{1}-k_{2}(x_{2}-x_{1})=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b58150ecdc2a496752002780f546f8cacd5afdd2)
![{\displaystyle m_{2}{\ddot {x}}_{2}=-k_{2}(x_{2}-x_{1})-k_{3}x_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0958df23d4b903b9992233dbafa9b48c9e99597)
- ∴
![{\displaystyle m_{2}{\ddot {x}}_{2}+k_{2}(x_{2}-x_{1})-k_{3}x_{2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7296e656c68e357e1c2b9526f2d0f255eb4fe09f)
where
and
represent the Newton's Second Law of Motion and
and
represent the net forces acting in the masses.
Laplace Transform
Applying the Laplace Transform to the motion equations and plugging the values of
,
,
,
,
,
,
,
, and
for this systems, we obtain,
![{\displaystyle X_{1}(s)(m_{1}s^{2}+k_{1}+k_{2})=m_{1}(sx_{1}(0)-{\dot {x}}_{1}(0))+k_{2}X_{2}(s)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7902464403f7a0a28a167a59ab3acff39a38bb2d)
![{\displaystyle X_{1}(s)={\dfrac {m_{1}(sx_{1}(0)+{\dot {x}}_{1}(0))+k_{2}X_{2}(s)}{(m_{1}s^{2}+k_{1}+k_{2})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40fdd08af71b70a63646d0fe795a80789b88354c)
![{\displaystyle X_{1}(s)={\dfrac {1(s0+(-1)]+1X_{2}(s)}{(1s^{2}+1+1)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1569d4323a471f35f58c9deed7a415e1d26dfdea)
![{\displaystyle X_{1}(s)={\dfrac {X_{2}(s)-1}{(s^{2}+2)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e08fde81ae32550207d8bb31f108fb54f9717cc)
![{\displaystyle X_{2}(s)(m_{2}s^{2}+k_{2}+k_{3})=m_{2}(sx_{2}(0)-{\dot {x}}_{2}(0))+k_{2}X_{1}(s)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24106b44adb501cbd1716a5d398e7b014afddd2a)
![{\displaystyle X_{2}(s)={\dfrac {m_{2}(sx_{2}(0)+{\dot {x}}_{2}(0))+k_{2}X_{1}(s)}{(m_{2}s^{2}+k_{2}+k_{3})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f368bf66e6afcb08684bb63d558969ddb15139ee)
![{\displaystyle X_{2}(s)={\dfrac {1(s0+1)+1X_{1}(s)}{(1s^{2}+1+1)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fff71391bb986c122f079afc6511e0f8255bc4cb)
![{\displaystyle X_{2}(s)={\dfrac {X_{1}(s)+1}{(s^{2}+2)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7dfd63d46c5814eab250d711e61c110fe8896801)
Finally, solving for
and
yields,
![{\displaystyle X_{1}(s)={\dfrac {-1}{s^{2}+3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2a6530a28f23453350c5e11cc1ef9fda2c1bc27)
![{\displaystyle X_{2}(s)={\dfrac {1}{s^{2}+3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bb8c61af4fff8ba1fc6741128b33224ae6f0911)
Inverse Laplace Transform
Figure 2. Coupled Spring System Motion.
First, we recognize that
On the other hand, we identify that
, and so
. Hence, we fix the expression by multiplying and dividing by
,
![{\displaystyle x_{1}(t)={\mathcal {L}}^{-1}[X_{1}(s)]={\mathcal {L}}^{-1}\left[{\dfrac {-1}{s^{2}+3}}\right]=-{\dfrac {1}{\sqrt {3}}}{\mathcal {L}}^{-1}\left[{\dfrac {\sqrt {3}}{s^{2}+3}}\right]=-{\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a32d68f215f452efdbcce10a11e1b233a8fc230)
![{\displaystyle x_{2}=(t){\mathcal {L}}^{-1}[X_{2}(s)]={\mathcal {L}}^{-1}\left[{\dfrac {1}{s^{2}+3}}\right]={\dfrac {1}{\sqrt {3}}}{\mathcal {L}}^{-1}\left[{\dfrac {\sqrt {3}}{s^{2}+3}}\right]={\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19364b5f445eb5a82d33a1d0b3ae49ec90388a72)
A plot of the oscillatory motion is shown on Figure 2.
Initial-Value & Final-Value Theorem
The initial-value and final-value theorem can be useful the finding the behavior of a function at small and large times respectively. By definition, the Initial-Value Theorem is,
and the Final-Value Theorem is,
Thus, applying both theorems to our the Laplace Transforms,
![{\displaystyle \lim _{s\rightarrow \infty }sX_{1}(s)=\lim _{s\rightarrow \infty }s{\dfrac {-1}{s^{2}+3}}=\lim _{s\rightarrow \infty }{\dfrac {-s}{s^{2}+3}}={\dfrac {-\infty }{\infty ^{2}+3}}=0=x_{1}(0)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4533de2bb69238eb570a2856431e4e8c1edade69)
![{\displaystyle \lim _{s\rightarrow 0}sX_{1}(s)==\lim _{s\rightarrow 0}s{\dfrac {1}{s^{2}+3}}=\lim _{s\rightarrow 0}{\dfrac {s}{s^{2}+3}}={\dfrac {0}{0^{2}+3}}=0=x_{1}(\infty )\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec61e3723f07f6e7685b98942b12e05266f9bc7d)
![{\displaystyle \lim _{s\rightarrow \infty }sX_{2}(s)=\lim _{s\rightarrow \infty }s{\dfrac {1}{s^{2}+3}}=\lim _{s\rightarrow \infty }{\dfrac {s}{s^{2}+3}}={\dfrac {\infty }{\infty ^{2}+3}}=0=x_{2}(0)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d57c2197424c0178faed85b4104080075ce709aa)
![{\displaystyle \lim _{s\rightarrow 0}sX_{2}(s)==\lim _{s\rightarrow 0}s{\dfrac {1}{s^{2}+3}}=\lim _{s\rightarrow 0}{\dfrac {s}{s^{2}+3}}={\dfrac {0}{0^{2}+3}}=0=x_{2}(\infty )\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85144ec4614c0f8e542d86b1c8161ce92327f6fa)
Bode Plot
The Bode plot for the Transfer Functions
and
can be easily done using a program like Octave or MATLAB. The code is displayed below. From Figure 3 we may notice that the Amplitude vs. Frequency plot for both functions overlaps. The peak amplitude occurs at
seconds, as well as the phase switching as shown in the Phase vs. Frequency plot.
h1=tf([-1],[1 0 3]);
h2=tf([1],[1 0 3]);
bode(h1,'b',h2,'-.r');
legend('H_1(s)','H_2(s)')
grid on;
Magnitude Frequency Response
Considering the Transfer Functions
and
described in the Bode Plot section, we notice that there are no values for the s-variable that make
and
equal zero.
From the Bode Plot, we notice that the break point occurs at ≈ 1.73 seconds, so the
Convolution
By definition, the convolution of two functions is,
where
refers to the inverse Laplace Transform. Assuming that
is a dummy variable of integration, and
is the impulse function
, the convolution for our system is,
![{\displaystyle y_{1}(t)=\int _{0}^{t}{\delta (t-\tau )h_{1}(\tau )d\tau }=-{\dfrac {1}{\sqrt {3}}}\int _{0}^{t}{\delta (t-\tau )sin({\sqrt {3}}\tau )d\tau }=\left[-{\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}\tau )\right]_{\tau =t}=-{\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/baf9033459afc69ee620fa1debe1275fe9444b2f)
![{\displaystyle y_{2}(t)=\int _{0}^{t}{\delta (t-\tau )h_{2}(\tau )d\tau }={\dfrac {1}{\sqrt {3}}}\int _{0}^{t}{\delta (t-\tau )sin({\sqrt {3}}\tau )d\tau }=\left[{\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}\tau )\right]_{\tau =t}={\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbc07a8904ce6a148b87065cac3256405797c0f1)
which is the same solution yields by the inverse Laplace Transform
State Equation Model
By definition, the the state equation is stated as
Now, consider the motion equations described in the Solution section,
Solving for
and
yields,
![{\displaystyle {\ddot {x}}_{1}=-{\dfrac {k_{1}}{m_{1}}}x_{1}+{\dfrac {k_{2}}{m_{1}}}x_{2}-{\dfrac {k_{2}}{m_{1}}}x_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00180f99227e5f96d594c6bebae044c465b88521)
![{\displaystyle {\ddot {x}}_{2}={\dfrac {-k_{2}}{m_{2}}}x_{2}+{\dfrac {k_{2}}{m_{2}}}x_{1}+{\dfrac {k_{3}}{m_{2}}}x_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/896d8a57b7acb143add5c78ac6d61b15de7ebf5c)
Finally, we let
,
,
, and
be the state variables. Thus,