A springy pad

Consider the system that consist of mass and spring.When system doesn't move it reachs equilibrium states,so that equation of motion is k \times h = m \times g then we obtain the ratio of \frac {k}{m} which is equal to: \frac {g}{h}...(1) where h is the compression of the spring Now from oscillatory motion of the spring-mass system we obtain: f = \frac {1}{2pi} \sqrt \frac {k}{m} ...(2) From equation (1) to (2) we obtain: f=\frac {1}{2pi} \sqrt \frac {g}{h} ...(3) and the unit system of this equation is revolution per second so we need to change second to minute so that equation (3) becomes f=\frac {60}{2pi} \sqrt \frac {g}{h} by substitute the value of g and h we get the rpm is 94.533

We will have the maximum vertical oscillations when the motor and the pad will be in resonance condition The resonant frequency of a spring and mass system is $\sqrt\frac{k}{m}$ $\frac{rad}{second}$

Also we have mg = kh hence $\frac{k}{m}$ =98 $\frac{rad}{m}$ Implies Rotational Speed $\frac{60}{2pi}$ $\sqrt{98}$ $\frac{revolutions}{min}$ = 94.53 $\frac{revolutions}{min}$

Please refer to the diagram above. In a spring mass system we have Kx=mg . Therefore, K/m=g/x. where K= the elastic co-efficient of the rubber pad. m=mass of the motor. g= The gravitational acceleration=9.8m./ sec2. x=10 cm. =0.1 m. The natural frequency of the system is ω=sqrt (K/m)=sqrt(g/x). the motor will exhibit the largest vertical vibration at resonance F= ω/(2 pi)= sqrt (K/m)/(2 pi) =sqrt(g/x)/(2 pi) =sqrt(9.8/0.1)/(2 3.14) rev. per sec. =sqrt(9.8/0.1) 60/(2 3.14) RPM.=94.53 RPM.

Let the elastic coefficient of the rubber pad be $k$ . Then $kx = mg$ , where $m$ is the mass of the motor. When $x = 0.1~\mbox{m}$ , the natural frequency of the system is:

$f_n=\frac1{2\pi}\omega_n=\frac1{2\pi}\sqrt{k/m} =\frac1{2\pi} \sqrt{g/x}= 1.58~\mbox{Hz}$

The motor will exhibit the largest vertical vibration when we have resonance, so the rotation frequency of the motor must also be equal to $f_n$ , which means the rotation speed is $60 \times f_n = 94.53~\mbox{RPM}$ .

Rotation per minute:1/2 phi 60