Prove that for finding the natural frequency of spring mass system, the mass of the spring can be taken in to account by adding one third of its mass of the main mass.

OR

P.T. $w_n = \sqrt{ \frac{k}{m + \frac{ms}{3}}}$

$\frac{\delta}{y} = \frac{x}{L}$

$\delta = (\frac{y}{L}) x$

$\delta' = (\frac{y}{L}) x'$

$Mass = \rho \times (volume)$

$= \rho \times [ A \times L]$

$M = \rho \times L$

$dM = \rho dy$

KE = (KE)m + (KE)dm

$\frac{1}{2} m x^2 + \int^L_0 \frac{1}{2} dM. \delta '^2$

$= \frac{1}{2} mx^2 + \frac{1}{2} \int^L_0 \rho . dy (\frac{y}{L} x)^2$

$= \frac{1}{2} mx^2 + \frac{1}{2} .\rho. \frac{x^2}{L^2} \int^L_0 y^2.dy$

$KE = \frac{1}{2} mx^2 + \frac{1}{2} \rho . \frac{x^2}{L^2} [ \frac{y^3}{3} ] ^L_0$

$= \frac{1}{2} m x^2 + \frac{1}{2} . \rho . \frac{x^2}{L^2} [ \frac{L^3}{3} ] $

$= \frac{1}{2} m x^2 + v [ [ \frac{\rho L}{3}] x^2$

$= \frac{1}{2} m. x^2 + \frac{1}{2} [ \frac{M}{3}] x^2$

$KE = \frac{1}{2} [ m + \frac{m}{3} ] x^2$

$M_{eg} = m + \frac{M}{3}$

$PE = \frac{1}{2} k. x^2$

$K_{eq} = K$

$w_n = \sqrt{ \frac{k}{m + \frac{m}{3}}}$

Diagram

Spring 1, 2 are in parallel

$\therefore K_{eq} 1,2 = K +K = 2K$

Above spring are in series.

$\therefore$ $\frac{1}{K_{eq}} = \frac{1}{2k} + \frac{1}{k} = \frac{1+2}{2k} = \frac{3}{2k}$

$Keq_{123} = \frac{2k}{3}$

$\therefore Keq = Keq_{123} cos^2_\alpha$

$= \frac{2k}{3} cos^2_\alpha$

$\therefore$ $w_n = \sqrt{ \frac{2k cos^2 \alpha}{3m}}$

Generally, m_{eq} x + k_{eq} x = 0

$m.x + \frac{2}{3} k cos^2 \alpha = 0$ - - - - equation of motion.

$\frac{k}{k_x} = \frac{1}{2k} + \frac{1}{2k} + \frac{1}{2k} + \frac{1}{2k cos^2 45}$

$= 0.5k +0.5k + 0.5k + \frac{0.5k}{cos^2 45}$

$= 0.5k (1+1+1+ \frac{1}{cos^2 45}$

$= 0.5k (3+ \frac{1}{cos^2 45})$

page 32:

$Keq_{5,6} = k + k = 2k$

$\frac{1}{keq_{4,5,6,7}} = \frac{1}{2k} + \frac{1}{2k} + \frac{1}{2k} = \frac{3}{2k}$

$Keq_{4-7} = \frac{2k}{3}$

$Keq_8 = 2k.cos^2 45$

$\frac{1}{Keq_{4,8}} = \frac{1}{\frac{2k}{3}} + \frac{1}{2k.cos^2 45}$

$Keq_{4-8} = \frac{2k}{3} + 2k.cos^2 45$

$Keq_{1-8} = k + k.cos^230 + \frac{2k}{3} + 2k. cos^2 45$

$= K(1 + cos^2 30) + 2k (\frac{1}{3} + cos^2 45)$

= 1.75 k + k = 2.75 k

$K_{eq} = 2.15k$

$w_n = \sqrt{ \frac{k_{eq}}{m_{eq}}}$

$w_n \sqrt{ \frac{2.15k}{m}} rad/sec$

k + 0.75k = 1.75k + 0.4k = 2.15k

page 33:

$(k_1 + k_2) 0.5 + (k_1 + k_2) 0.5$

$0.5k_1 + 0.5k_2 + 0.5k_1 + 0.5k_2$

$K_{eq} = k_q + k_2$

$w_n = \sqrt{ \frac{k_1 + k_2}{m}} rad/s$

$K_{eq} = (k_2 + k_1) cos^2 45$

$= 0.5 (k_1 + k_1)$

$K_{eq} = 0.5 (k_1 + k_2) + 0.5 (k_1 + k_2)$

$= (k_1 + k_2)$

$K_{eq} = (k_1 + k_2) cos^2 45$

$= 0.5 (k_1 + k_2)$