Coupled oscillators

Let $x_1$ be the displacement of $m_1$ from its equilibrium position, $x_2$ be the displacement of $m_2$ from its equilibrium position, $\dot x = \frac{\mathrm dx}{\mathrm dt}, \ddot x = \frac{\mathrm d^2 x}{\mathrm d t^2}$ . By using Hooke’s Law, we can have two equations of motions:

$m \ddot x_1= k(-2x_1 + x_2) \implies \ddot x_1=-32x_1 + 16x_2$ $m \ddot x_2 = k(x_1 - 2x_2) \implies \ddot x_2 = 16x_1 - 32x_2$

Method 1:

By rearranging the equations above, we can get: $\ddot x_2 = -48x_1-2\ddot x_1$

Differentiate the $\ddot x_1=-32x_1 + 16x_2$ w.r.t $t$ , we get:

$x_1^{(4)}= -32\ddot x_1 + 16\ddot x_2= -32\ddot x_1 + 16(-48x_1 - 2\ddot x_1)$ $\implies x_1^{(4)} + 64 \ddot x_1 + 768x_1=0$

By solving the characteristic equations like those of the second order linear homogeneous differential equations, we get four roots. The general equation is $x_1= c_1 \sin 4t + c_2 \cos 4t + c_3 \sin \sqrt{48} t + c_4 \cos \sqrt{48} t$ .

Sub this equation into $\ddot x_1=-32x_1 + 16x_2$ , we get:

$x_2= c_1 \sin 4t + c_2 \cos 4t - c_3 \sin \sqrt{48} t - c_4 \cos \sqrt{48} t$ .

The initial conditions are: $x_1(0)= 10, \dot x_1(0)=0, x_2(0)=0, \dot x_2(0)=0$

When we use the initial conditions, we get $c_1=0, c_2=5, c_3=0, c_4=5$

$x_2= 5\cos 4t - 5\cos \sqrt{48} t$

When $t=5$ , $x_2=7.023 cm$

Method 2:

The two simultaneous differential equations can be solved by using eigenvector method.

$\begin{pmatrix} \ddot x_1 \\ \ddot x_2 \\ \end{pmatrix} = \begin{pmatrix} -32 & 16\\ 16 & -32 \end{pmatrix}\begin{pmatrix} x_1\\ x_2\end{pmatrix}$

There are two eigenvalues, $\lambda_1= -16$ and $\lambda_2=-48$ . For $\lambda_1$ , its eigenvector is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ . For $\lambda_2$ , its eigenvector is $\begin{pmatrix} 1\\ -1 \end{pmatrix}$ .

Thus the general solution is $\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} 1\\1 \end{pmatrix} (c_1 \sin 4t + c_2 \cos 4t) + \begin{pmatrix} 1\\-1 \end{pmatrix} (c_3 \sin 4t + c_4 \cos 4t)$

When we sub in the initial conditions, we get the same answer: at $t=5$ , $x_2= 7.023 cm$ .

Method 3:

The coupled oscillators actually have two normal modes, and the $x_1$ and $x_2$ consists of the superposition of the two normal modes. One normal mode is when the two masses move in phase, i.e. the middle spring always keeps its original length (neither stretched nor compressed). In this case, $x_1=x_2$ . When we sub this equation into $\ddot x_1=-32x_1 + 16x_2$ , we get $\ddot x_1= -16 x_1$ . This is the equation of simple harmonic motion, so in the first normal mode, $\omega=4$ .

The second normal mode corresponds to the situation where two masses are $\pi$ radians out of phase, i.e. $x_1=-x_2$ . When we sub this condition into the $\ddot x_1=-32x_1 + 16x_2$ , we get $\ddot x_1 = -48x_1$ and $\omega= \sqrt{48}$ .

Hence $x_1$ is the superposition of the two modes and the general equation is $x_1= c_1 \sin 4t + c_2 \cos 4t + c_3 \sin \sqrt{48} t + c_4 \cos \sqrt{48} t$ . Since $x_1=x_2$ for $\omega=4$ and $x_1=-x_2$ for $\omega= \sqrt{48}$ ,

$x_2= c_1 \sin 4t + c_2 \cos 4t - c_3 \sin \sqrt{48} t - c_4 \cos \sqrt{48} t$

When we sub in the initial conditions, we get the answer: at $t=5$ , $x_2=7.023 cm$

Let write the equations of motions as below: $m\frac{d^2x_{1}}{dt^2} + m\omega^2_0x_{1} + k(x_{1} - x_{2}) = 0$ $m\frac{d^2x_{2}}{dt^2} + m\omega^2_0x_{2} - k(x_{1} - x_{2}) = 0$ Where $\omega_{0}$ is the natural frequency of either of the spring. By put $\omega_{c} = \sqrt{\frac{k}{m}}$ (Hooke’s Law) hence we get the new equation as below.

$\frac{d^2x_{1}}{dt^2} + (\omega^2_0 + \omega^2_c)x_{1} - \omega^2_cx_{2} = 0$ $\frac{d^2x_{2}}{dt^2} + (\omega^2_0 + \omega^2_c)x_{2} - \omega^2_cx_{1} = 0$ We can rewrite the equations: $\frac{d^2q_{1}}{dt^2} + \omega^2_0q_{1} = 0$ $\frac{d^2q_{2}}{dt^2} + \omega'^2q_{2} = 0$ With $q_{1} = Acos(\omega_{0}t)$ and $q_{2} = Bcos(\omega’^2t).$ Thus we get $x_{1} = q_{1} + q_{2} = Acos(\omega_{0}t) + Bcos(\omega’t)$ $x_{2} = q_{1} + q_{2} = Acos(\omega_{0}t) - Bcos(\omega’t)$ With $\omega_0 = \sqrt{\frac{k}{m}} = 4$ and $\omega’ = \sqrt{\frac{3k}{m}} = 4\sqrt{3}$

By solving when it in initial condition we may obtain $A = B = 5$ cm

Hence, the absolute value of the distance from the equilibrium position when $t =5$ is

$= |5\cos(5\omega_{1}) - 5\cos(5\omega_{2})|$

$= |5\cos(5 \times 4) - 5\cos(5 \times 4\sqrt{3})|$

$= 7.023$ cm

According to Newton Law II, we have:

$mx''_1=-kx_1-k(x_1-x_2)$

and $mx''_2=-kx_2+k(x_1-x_2)$ .

$$

Therefore, $m(x_1+x_2)''=-k(x_1+x_2)$

and $m(x_1-x_2)''=-3k(x_1-x_2)$ .

$$

Therefore, $x_1+x_2=A \cos (\omega_1 t+\phi_1)$

and $x_1-x_2=B \cos (\omega_2 t+\phi_2)$ , with $\omega_1=\sqrt{\frac{k}{m}}, \omega_2=\sqrt{\frac{3k}{m}}$ .

$$

When t=0, $x_1=d, x_2=0,x'_1=x'_2=0$ , so $A=B=d, \phi_1=\phi_2=0$ .

Therefore, $x_2=\frac{d}{2} (\cos \omega_1 t - \cos \omega_2 t)$ .

Therefore, when $t=t'=5s, x_2=7.02 cm$

By symmetry, the two modes of the problem are the symmetric mode ( $x_2=x_1$ ) and the antisymmetric mode ( $x_2=-x_1$ ).

In the symmetric mode, the spring between the blocks is always at equillibrium length. So the force the first block feels is only from first spring and the time period is same as that for a single block-spring system : $\omega_1=\sqrt{\frac{k}{m}}$

In the antisymmetric mode, the elongation of the central spring is twice that of either end spring. Hence the block feels thrice as much force as in a single block-spring system : $\omega_2=\sqrt{\frac{3k}{m}}$

Since both blocks start from rest, displacements of the two blocks can be written as: $x_1=A\cos(\omega_1t)+B\cos(\omega_2t)$ $x_2=A\cos(\omega_1t)-B\cos(\omega_2t)$

Solving for initial conditions gives $A=B=5\;\rm{cm}$

Hence the required answer is $5(\cos(5 \omega_1)-\cos(5 \omega_2))=7.023\;\rm{cm}$

PS: If someone is not convinced of the symmetry argument for the normal modes, write out the total force in terms of displacements of either block. You will find that the governing equations reduce to SHM equations with change of variables $y=x_1+x_2,z=x_1-x_2$

When the $m_1$ block moves a distance $d= 10 cm$ towards right.

Then one can easily obtained the equation of motions from their respective free body diagrams of $m_1$ amd $m_2$ blocks .

BUt given that the masses are same . Let us say m 1 = m 2 = m) .

Here are the equations

$mx_1'' + kx_1 - k(x_2 - x_1) = 0$

$mx_2'' + kx_2 - k(x_1 - x_2) = 0$

Rearranging these two equations and dividing both two equations by $m$ . we obtain ,

$x_1'' + (\frac{2k}{m})x_1 - (\frac{k}{m})x_2 = 0$

$x_2'' + (\frac{2k}{m})x_2 - (\frac{k}{m})x_1 = 0$

we first add the two equations together, and then subtract them, to obtain two new equations

$(x_1 + x_2)'' + (\frac{k}{m})(x_1 + x_2) = 0$ - (1)

$(x_1 - x_2)'' + (\frac{3k}{m})(x_1 - x_2) = 0$ - (2)

Now we change to our normal coordinates $q_1$ and $q_2$ . We will also make a substitution for $w_1$ and $w_2$ , the normal frequencies of the system. It turns out that these normal frequencies are the frequencies at which the masses oscillate in their normal modes of vibration.

$q_1 = x_1 + x_2$ and $w_1 = \frac{k}{m}$

$q_2 = x_1 - x_2$ and $w_2 = \frac{3k}{m}$

Substituting into equation (1) and (2) , we obtain ;

$q_1'' +w_1^2q_1 = 0$

$q_2'' + w_2^2q_2 = 0$

NOTE : THIS DIFFERNTIAL EQUATION IS SIMILAR TO HARMONIC OSCILLATOR. We all know the solution of such kind of second order differential equation , which is of the form

$q_1(t)= c_1cos(w_1t) + c_2sin(w_1t)$

$q_2(t)= c_3cos(w_2t) + c_4sin(w_2t)$

Now we use the relationship $x_1 = \frac{q_1 + q_2}{2}$

$x_2 = \frac{q_1 - q_2}{2}$

to give the general solution for $x_1$ and $x_2$ .

$x_1(t) = c_1cos(w_1t) + c_2sin(w_1t) + c_3cos(w_2t) + c_4sin(w_2t)$ -(3)

$x_2(t) = c_1cos(w_1t) + c_2sin(w_1t) - c_3cos(w_2t) - c_4sin(w_2t)$ -(4)

In the ques , we are given that the block $m_1$ moved towards right a distance $d = 10 cm$ .

Thus ,

$x_1(0) = d$ and $x_1'(0) = 0$

$x_2(0) = 0$ and $x_2'(0) = 0$

After substituting these values into Equation 3 and Equation 4 we can solve for the constants $c_1, c_2, c_3, and c_4$ .

Hence , we have our final solutions ,

$x_1 = \frac{d}{2}[ cos(w_1t) + cos(w_2)t ]$ - (5)

$x_2 = \frac{d}{2}[ cos(w_1t) - cos(w_2t) ]$ - (6)

Hence in equation (6) . Put , the values of $w_1$ and $w_2$ and $t=5$ , then solve for $x_2$ .

WE will get $x_2 = 7.023 cm$ .

Let's write the moviment equations for $x_1$ and $x_2$ : $$ m\ddot{x} 1 = k(x 2 - x 1) - kx 1 $$ $$ m\ddot{x} 2 = -k(x 1 - x 2) - kx 2 $$ Leading us to the following system: $$ \ddot{x} 1 = -\frac{2k}{m}x 1 + \frac{k}{m}x 2 $$ $$ \ddot{x} 2 = \frac{k}{m}x 1 - \frac{2k}{m}x 2 $$ Finding the eigenvalues of the coefficients matrix: $$ \lambda 1 = - \frac{k}{m} \quad \lambda 2 = -\frac{3k}{m} $$ Which represents the normal modes frequencies. The eigenvectors for these eigenvalues are respectively $\nu_1 = (1 \quad 1)^T$ and $\nu_2 = ( 1 \quad -1)^T$ . Let's call $M = (\nu_1 \quad \nu_2)$ .$$ $$Creating $Y = MX$ where $X = (x_1 \quad x_2)^T$ we'll get the following: $$ Y = MX \Rightarrow \ddot{Y} = M\ddot{X} $$ $$ \ddot{X} = AX \Rightarrow M\ddot{Y} = AMY \Rightarrow \ddot{Y} = M^{-1}AMY \Rightarrow \ddot{Y} = HY $$ Where $A$ is the coefficients matrix and $H$ the eigenvalues matrix. Solving the new system for $y_i$ we'll have: $$ \ddot{y} 1 = -\frac{k}{m}y 1 \Rightarrow \ddot{y} 1 + \frac{k}{m}y 1 = 0 $$ $$ \ddot{y} 2 = -\frac{3k}{m}y 2 \Rightarrow \ddot{y} 2 + \frac{3k}{m}y 2 = 0 $$ Which are clearly simple harmonic motions with frequencies $\omega_1 = \sqrt{\frac{k}{m}} \quad \omega_2 = \sqrt{\frac{3k}{m}}$ . Solving theses equations we have: $$ y 1 = c 1 \cos \left( \omega 1t + \phi 1\right), \quad y 2 = c 2 \cos \left( \omega 2t + \phi 2 \right) $$ Now, getting $x_1$ and $x_2$ : $$ Y = MX \Rightarrow X = M^{-1}Y $$ We'll find: $$ x 1 = k 1\cos(\omega 1t + \phi 1) + k 2\cos(\omega 2t + \phi 2)
$$ $$ x
2 = k 1\cos(\omega 1t + \phi 1) - k 2\cos(\omega 2t + \phi 2) $$ Where $k_i = \frac{c_1}{2}$ . Applying the initial conditions we'll find: $$ k 1 = k 2 = \frac{d}{2}, \quad \phi 1 = \phi 2 = 0 $$ So, let's find $x_2(5)$ : $$ x 2(t) = \frac{d}{2}\cdot\cos\left(\sqrt{\frac{k}{m}}\cdot t + 0 \right) - \frac{d}{2}\cdot\cos\left(\sqrt{\frac{3k}{m}}\cdot t + 0 \right) $$ $$ x 2(5) = 5(\cos(20) - \cos(20\cdot\sqrt{3})) \Rightarrow x_2(5) = 7.023 $$

Assume that the positive direction is rightward and the displacements of m {1}, m {2} are x {1},x {2} respectively. The derivative equation of motion of two objects; mx {1}'' = -kx {1} + k(x {2}-x {1}) => mx {1}'' + 2*kx {1} - kx {2} =0 (1) mx {2}'' = -kx {2} + k(x {1}-x {2}) => mx {2}'' + 2 kx_{2} - kx_{1} =0 (2) From (1) +(2) : m(x_{1} +x_{2})'' + k(x_{1}+x_{2}) = 0 at t=0, (x_{1} +x_{2})'=0 therefore x_{1} +x_{2}= (10+0) cos (\omega t) where 10,0 are the initial values of x {1} and x {2}; \omega= \sqrt{\frac{k}{m}} = 4 From ( 1) - (2) we get: m(x {1} -x {2})'' +3 k(x_{1}-x_{2}) = 0 at = 0: (x_{1} -x_{2})'=0 therefore x_{1} -x_{2} = 10 cos (4sqrt{3} t) Substitute t=5, we can figure out x {1} + x {2}, and x {1} -x {2} then figure out x_{2} = 7.023