**1 Answer**

written 4.9 years ago by |

**Show that the frequency of vibration is minimum when the load point is at the centre of fixed ends (Y = 0.5L)**

where,

y = Dist between fixed end and load point.

L = Dist. between the fixed ends.

**Case 1:** When a helical spring is considered.

**Case 2:** When mass is attached between two helical springs.

**Case 1 -** k = $\frac{Gd^4}{8D^3n}$

Let, L = length of spring

n = no of turns.

$\rightarrow$ $L = \pi D.n$

$\therefore$ $n = \frac{L}{\pi D}$

$k = \frac{G.d^4}{8D^3 (\frac{L}{\pi D})}$

All the terms are constant except 'L'

$K \alpha \frac{1}{L}$

or $K = \frac{C}{L}$

C = KL - - - - (1)

**Case 2:** $k_1 = \frac{c}{y}$

$k_2 = \frac{c}{L-Y}$

since $k_1$ and $k_2$ are in parallel

$Keq = [ \frac{c}{y} + \frac{c}{L-Y} ]$

$w_n = \sqrt{ \frac{Keq}{m}} = \sqrt{ \frac{1}{m} [ \frac{c}{y} + \frac{c}{L-Y}]}$

$w_n = \sqrt{ \frac{KL}{m} [ \frac{1}{y} + \frac{1}{L-Y}]} $

$\frac{d}{dy} [ \frac{K_L}{m} ( \frac{1}{y} + \frac{1}{L-Y})]^1/2 = 0$

$(\frac{K_L}{m}^1/2 \frac{d}{dy} [ \frac{1}{4} + \frac{1}{L-Y} ]^1/2 = 0$

$\frac{d}{dy} [ \frac{1}{4} + \frac{1}{L-4}]^1/2 = 0$

$\frac{1}{2} [ \frac{-1}{y^2} + \frac{1}{(L-Y)^2}] = 0$

$\frac{1}{(L-Y)^2} = \frac{1}{y^2}$

Y = L - Y

L = 2Y

$\therefore Y = 0.5L$