Question 17.4: A mass of 10 kg is fixed in the middle of a spring of stiffn...

Deflection of a helical spring under axial load is:

\delta=\frac{8 W D_{m}^{3} n}{G d^{4}} .

∴          \text { Stiffness, } k=\frac{W}{\delta}=\frac{G d^{4}}{8 D_{m}^{3} n} .

or      k \propto \frac{1}{n} .

when the mass m = 10 kg is fixed in the middle of the spring, the number of turns in each portion becomes n /2. Thus stiffness of each portion becomes 2k. Since the two portions of the spring are in parallel, therefore total stiffness becomes 4k.

Natural frequency,        \omega_{n}=\sqrt{\frac{4 k}{m}} .

=\sqrt{\frac{4 \times 5 \times 10^{3}}{10}} .

=44.72 rad / s .

or        f_{n}=\frac{\omega_{n}}{2 \pi}=\frac{44.72}{2 \pi}=7.12 Hz .