A body of mass 20 kg is suspended from a spring which deflects 15 mm under this load. Calculate the frequency of free vibrations and verify that a viscous damping force amounting to approximately 1000 N at a speed of 1 m/s is just-sufficient to make the motion aperiodic.
If when damped to this extent, the body is subjected to a disturbing force with a maximum value of 125 N making 8 cycles/s, find the amplitude of the ultimate motion.
Question : A body of mass 20 kg is suspended from a spring which deflec...
Given : m = 20 kg ; \delta=15 mm =0.015 m ; c = 1000 N/m/s ; F = 125 N ; f = 8 cycles/s
Frequency of free vibrations
We know that frequency of free vibrations,
f_{n}=\frac{1}{2 \pi} \sqrt{\frac{g}{\delta}}=\frac{1}{2 \pi} \sqrt{\frac{9.81}{0.015}}=4.07 HzThe critical damping to make the motion aperiodic is such that damped frequency is zero, i.e
\left(\frac{c}{2 m}\right)^{2}=\frac{s}{m}
\therefore c=\sqrt{\frac{s}{m} \times 4 m^{2}}=\sqrt{4 s . m}=\sqrt{4 \times \frac{m \cdot g}{\delta} \times m} \ldots\left(\because s=\frac{m \cdot g}{\delta}\right)
=\sqrt{4 \times \frac{20 \times 9.81}{0.015} \times 20}=1023 N / m / s
This means that the viscous damping force is 1023 N at a speed of 1 m/s. Therefore a viscous damping force amounting to approximately 1000 N at a speed of 1 m/s is just sufficient to make the motion aperiodic.
Amplitude of ultimate motion
We know that angular speed of forced vibration,
\omega=2 \pi \times f=2 \pi \times 8=50.3 rad / sand stiffness of the spring, s=m \cdot g / \delta=20 \times 9.81 / 0.015=13.1 \times 10^{3} N / m
\therefore Amplitude of ultimate motion i.e. maximum amplitude of forced vibration,
x_{\max }=\frac{F}{\sqrt{c^{2} \cdot \omega^{2}+\left(s-m \cdot \omega^{2}\right)^{2}}}=\frac{125}{\sqrt{(1023)^{2}(50.3)^{2}+\left[13.1 \times 10^{3}-20(50.3)^{2}\right]^{2}}}
=\frac{125}{\sqrt{2600 \times 10^{6}+1406 \times 10^{6}}}=\frac{125}{63.7 \times 10^{3}}=1.96 \times 10^{-3} m
= 1.96 mm