## Transverse Vibrations of Three Particle Elastic String System With Forcing Terms

Three particles 1, 2 and 3 are fixed to three horizontal elastic strings, each of length
$l$
, and sibject to horizontal forces .
$p_1 sin (k_1t), \: p_2 sin (k_2t), \: p_3 sin (k_3t)$
.

Suppose the displacements are
$x_1, \: x_2$
and
$x_3$
respectively. If the increase
$x_1$
in the length of string 1 is small then we can treat
$T$
as constant. Resolving vertically for each particle in turn: Particle 1:
$m_1 \ddot{x}_1= -Tsin \theta_1 +T sin \theta_2+ p_1 sin (k_1t)$

Particle 2:
$m_2 \ddot{x}_2= -Tsin \theta_2 -T sin \theta_3+ p_2 sin (k_2t)$

Particle 2:
$m_3 \ddot{x}_3= Tsin \theta_3 -T sin \theta_4+ p_3 sin (k_3t)$

For each
$x$
small
$tan \theta =\frac{x}{l} \simeq \theta$
so the above equations become
Particle 1:
$m_1 \ddot{x}_1= -\frac{Tx_1}{l} +\frac{T(x_2-x_1}{l}+ p_1 sin (k_1t)$

Particle 2:
$m_2 \ddot{x}_2= -\frac{T(x_2-x_1)}{l} -\frac{Tx_2-x_3)}+p_2 sin (k_2t){l}$

Particle 2:
$m_3 \ddot{x}_3= T\frac{x_2-x_3}{l} - \frac{x_3}{l}+p_3 sin (k_3t)$

Dividing by
$m_1, \: m_2, \: m_3$
respectively and writing in matrix form gives
$\begin{pmatrix}\ddot{x}_1 \\ \ddot{x}_2 \\ \ddot{x}_3 \end{pmatrix} = \left( \begin{array}{ccc} -\frac{2T}{m_1l} & \frac{T}{m_1l} & 0 \\ \frac{T}{m_2l} & -\frac{2Tx_2}{m_2l} & \frac{T}{m_2l} \\ 0 & \frac{T}{m_3l} & - \frac{T}{m_3l} \end{array} \right) \begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix} + \begin{pmatrix}\frac{p_1}{m_1} sin (k_1t) \\ \frac{p_2}{m_2} sin (k_2t) \\ \frac{p_3}{m_3} sin (k_3t) \end{pmatrix}$

The eigenvalues of the matrix are
$\lambda$
and natural frequencies of vibration are
$f=2 \pi \sqrt{- \lambda}$
.
If any od the forcing frequencies
$f=_i- 2 \pi k_i$
are equal to the corresponding natural frequencies then we have resonance.