Oscillations and Waves > The superposition of oscillations > Problem 3.4.15

Statement

$3.4.15.$ a. Two balls of mass $m$, which are connected to each other and to the walls by three identical springs of stiffness $k$, are simultaneously given the same modulo velocity directed along the springs. Find the frequency of oscillation of the balls if their velocities are oppositely directed / equally directed.

b. Free vibrations of complex systems are the sum (superposition) of several harmonic oscillations with different frequencies. If the first ball in Problem 3.4.15a is given a velocity $v$ along the spring, then the subsequent motion of the balls will be the sum of two motions: the motion of the balls given a velocity $v/2$ and $-v/2$, and the motion of the balls given a velocity $v/2$ and $v/2$. Use this to determine the velocity of the balls at the moments following the beginning of the oscillations. What is the maximum displacement of the first ball? the second ball? the maximum elongation of the middle spring?

c. Solve the problem 3.4.15b if the first ball has a velocity $3v$ and the second one has a velocity $v$.

Solution

Part a: Normal Modes and Frequencies

Let $x_1$ and $x_2$ be the displacements of the first and second balls from their respective equilibrium positions. The system consists of three springs: the left spring is stretched by $x_1$, the right spring is compressed by $x_2$, and the middle spring is stretched by $(x_2 - x_1)$.

The equations of motion for the two masses are:
$$m \ddot{x}_1 = -kx_1 + k(x_2 - x_1) = -2kx_1 + kx_2$$
$$m \ddot{x}_2 = -kx_2 - k(x_2 - x_1) = kx_1 - 2kx_2$$

We analyze the two specific initial velocity configurations to find the natural normal mode frequencies:

1. **Oppositely Directed Velocities (Symmetric Mode):**
   Here, the balls move toward or away from each other symmetrically, meaning $x_1(t) = -x_2(t)$. Substituting $x_2 = -x_1$ into the first equation of motion yields:
   $$m \ddot{x}_1 = -2kx_1 + k(-x_1) = -3kx_1 \implies \ddot{x}_1 + \frac{3k}{m}x_1 = 0$$
   Thus, the angular frequency for oppositely directed initial velocities is:
   $$\omega_1 = \sqrt{\frac{3k}{m}}$$

2. **Equally Directed Velocities (Antisymmetric Mode):**
   Here, the balls move together in the same direction with identical displacements, meaning $x_1(t) = x_2(t)$. The middle spring undergoes no deformation ($(x_2 - x_1) = 0$). Substituting $x_2 = x_1$ into the first equation of motion yields:
   $$m \ddot{x}_1 = -2kx_1 + k(x_1) = -kx_1 \implies \ddot{x}_1 + \frac{k}{m}x_1 = 0$$
   Thus, the angular frequency for equally directed initial velocities is:
   $$\omega_2 = \sqrt{\frac{k}{m}}$$

Part b: Superposition of Initial Velocity $v$ on the First Ball

We decompose the initial state where $v_1(0) = v$ and $v_2(0) = 0$ into a linear combination of the two normal modes found in Part a:
* **Mode 1 (Symmetric):** Initial velocities are $v/2$ for ball 1 and $-v/2$ for ball 2. The velocity solutions are:
$$v_{1I}(t) = \frac{v}{2}\cos(\omega_1 t), \quad v_{2I}(t) = -\frac{v}{2}\cos(\omega_1 t)$$
The displacement solutions (integrating with $x(0)=0$) are:
$$x_{1I}(t) = \frac{v}{2\omega_1}\sin(\omega_1 t), \quad x_{2I}(t) = -\frac{v}{2\omega_1}\sin(\omega_1 t)$$

* **Mode 2 (Antisymmetric):** Initial velocities are $v/2$ for ball 1 and $v/2$ for ball 2. The velocity solutions are:
$$v_{1II}(t) = \frac{v}{2}\cos(\omega_2 t), \quad v_{2II}(t) = \frac{v}{2}\cos(\omega_2 t)$$
The displacement solutions are:
$$x_{1II}(t) = \frac{v}{2\omega_2}\sin(\omega_2 t), \quad x_{2II}(t) = \frac{v}{2\omega_2}\sin(\omega_2 t)$$

Superimposing the two modes ($x = x_I + x_{II}$ and $v = v_I + v_{II}$), we obtain the instantaneous velocities as a function of time:
$$v_1(t) = \frac{v}{2}(\cos \omega_1 t + \cos \omega_2 t)$$
$$v_2(t) = \frac{v}{2}(\cos \omega_2 t - \cos \omega_1 t)$$

The instantaneous displacements are:
$$x_1(t) = \frac{v}{2\omega_1}\sin \omega_1 t + \frac{v}{2\omega_2}\sin \omega_2 t$$
$$x_2(t) = -\frac{v}{2\omega_1}\sin \omega_1 t + \frac{v}{2\omega_2}\sin \omega_2 t$$

Finding Maximum Values:

1. **Maximum displacement of the first ball ($x_{1,\max}$):**
   Over a long period, the two independent sine terms will eventually reach their peak value of $1$ simultaneously:
   $$x_{1,\max} = \frac{v}{2\omega_1} + \frac{v}{2\omega_2} = \frac{v}{2}\left(\sqrt{\frac{m}{3k}} + \sqrt{\frac{m}{k}}\right) = \frac{v}{2}\sqrt{\frac{m}{k}}\left(1 + \frac{1}{\sqrt{3}}\right)$$

2. **Maximum displacement of the second ball ($x_{2,\max}$):**
   Similarly, constructive peak alignment occurs when $\sin \omega_2 t = 1$ and $\sin \omega_1 t = -1$:
   $$x_{2,\max} = \frac{v}{2}\sqrt{\frac{m}{k}}\left(1 + \frac{1}{\sqrt{3}}\right)$$

3. **Maximum elongation of the middle spring ($\Delta x_{\text{mid},\max}$):**
   The elongation of the middle spring is given by $\Delta x_{\text{mid}}(t) = x_2(t) - x_1(t) = -\frac{v}{\omega_1}\sin \omega_1 t$. The maximum absolute value occurs when $|\sin \omega_1 t| = 1$:
   $$\Delta x_{\text{mid},\max} = \frac{v}{\omega_1} = v\sqrt{\frac{m}{3k}}$$

Part c: General Initial Velocities $v_1(0) = 3v$ and $v_2(0) = v$

We resolve the initial conditions into mode amplitudes $C_1$ (Symmetric) and $C_2$ (Antisymmetric):
$$v_{1I}(0) = C_1, \quad v_{2I}(0) = -C_1$$
$$v_{1II}(0) = C_2, \quad v_{2II}(0) = C_2$$

Setting up the initial value system:
$$v_1(0) = C_1 + C_2 = 3v$$
$$v_2(0) = -C_1 + C_2 = v$$

Adding and subtracting these two equations yields the mode coefficients:
$$2C_2 = 4v \implies C_2 = 2v$$
$$2C_1 = 2v \implies C_1 = v$$

Substituting these components into the harmonic time profiles gives the general velocity functions:
$$v_1(t) = v \cos \omega_1 t + 2v \cos \omega_2 t$$
$$v_2(t) = -v \cos \omega_1 t + 2v \cos \omega_2 t$$

Integrating the velocities yields the equations for displacement:
$$x_1(t) = \frac{v}{\omega_1}\sin \omega_1 t + \frac{2v}{\omega_2}\sin \omega_2 t$$
$$x_2(t) = -\frac{v}{\omega_1}\sin \omega_1 t + \frac{2v}{\omega_2}\sin \omega_2 t$$

Finding Maximum Values for the General Case:

1. **Maximum displacement of the first ball ($x_{1,\max}$):**
   $$x_{1,\max} = \frac{v}{\omega_1} + \frac{2v}{\omega_2} = v\sqrt{\frac{m}{3k}} + 2v\sqrt{\frac{m}{k}} = v\sqrt{\frac{m}{k}}\left(2 + \frac{1}{\sqrt{3}}\right)$$

2. **Maximum displacement of the second ball ($x_{2,\max}$):**
   $$x_{2,\max} = \frac{v}{\omega_1} + \frac{2v}{\omega_2} = v\sqrt{\frac{m}{k}}\left(2 + \frac{1}{\sqrt{3}}\right)$$

3. **Maximum elongation of the middle spring ($\Delta x_{\text{mid},\max}$):**
   The elongation function reduces to $\Delta x_{\text{mid}}(t) = x_2(t) - x_1(t) = -\frac{2v}{\omega_1}\sin \omega_1 t$. Its maximum value is:
   $$\Delta x_{\text{mid},\max} = \frac{2v}{\omega_1} = 2v\sqrt{\frac{m}{3k}}$$

Answer

$$\boxed{\text{Oppositely directed: } \omega_1 = \sqrt{\frac{3k}{m}}, \quad \text{Equally directed: } \omega_2 = \sqrt{\frac{k}{m}}}$$

$$\boxed{v_1(t) = \frac{v}{2}(\cos \omega_1 t + \cos \omega_2 t), \quad v_2(t) = \frac{v}{2}(\cos \omega_2 t - \cos \omega_1 t)}$$

$$\boxed{x_{1,\max} = x_{2,\max} = \frac{v}{2}\sqrt{\frac{m}{k}}\left(1 + \frac{1}{\sqrt{3}}\right), \quad \Delta x_{\text{mid},\max} = v\sqrt{\frac{m}{3k}}}$$

$$\boxed{v_1(t) = v \cos \omega_1 t + 2v \cos \omega_2 t, \quad v_2(t) = -v \cos \omega_1 t + 2v \cos \omega_2 t}$$

$$\boxed{x_{1,\max} = x_{2,\max} = v\sqrt{\frac{m}{k}}\left(2 + \frac{1}{\sqrt{3}}\right), \quad \Delta x_{\text{mid},\max} = 2v\sqrt{\frac{m}{3k}}}$$