Question 12.1: An electric motor operating at a constant speed of 1000 rpm ...

The basic equation of motion for the vertical motion of the system is

m\ddot{x} +b\dot{x} +kx=F_{m} ,     (12.14)

where force F_{m} is a sinusoidal function of time having amplitude 400 N and frequency  ω= 2π1000/60 = 104.7 rad/s,

F_{m} =400\sin (104.7t) .     (12.15)

The amplitude and phase of displacement x(t) can be found by use of Eqs. (12.8) and (12.9). The transfer function relating the displacement to force F_{m} is obtained from Eq. (12.14), assuming zero initial conditions, {x}(0)=0 and \dot{x}(0)=0:

T_{s} =\frac{X(s)}{F_{m}(s) } =\frac{1}{ms^{2}+bs+k } .      (12.16)

One obtains the frequency-response transfer function of the system by putting s =jω in Eq. (12.16):

{T(j\omega )}={T(s)\mid _{s} }=\frac{1}{-m\omega ^{2}+j\omega b+k } .      (12.17)

For the given values of the system parameters and the frequency of vibration, ω = 104.7 rad/s, the frequency response transfer function is

{T(j\omega)\mid _{\omega=104.7} }=\frac{1}{-224,52+j83,760 } =1.75\times 10^{-11} (-224,052-j83,760) .     (12.18)

Note that, in Eq. (12.18), the complex number was rationalized (eliminating the imaginary part from the denominator) by multiplying the numerator and denominator by the complex conjugate of the denominator. Hence the magnitude of the transfer function at the frequency of interest is

\left|T(j\omega)\right| =1.75\times 10^{-11} \sqrt{\left(−224,052\right) ^{2}+\left(−83,760\right) ^{2} } =4.18\times 10^{-6} m/N .     (12.19)

From Eq. (12.8), the amplitude of the vertical displacement is

\left|X(j\omega)\right| =\left|F_{m} (j\omega)\right|\left|T(j\omega)\right|=400\times 4.18\times 10^{-6}=1.67\times 10^{-3} m .      (12.20)

From Eq. (12.9), the phase angle of the vertical displacement is equal to the phase angle of the frequency response transfer function. Note that, from Eq. (12.18), it can be seen that this complex number is in the third quadrant (both imaginary and real parts are negative) and the proper trigonometric equation must be used:

\phi _{T} =-\left[\pi -\tan ^{-1}\left(\frac{−83,760}{−224,052} \right) \right] =-2.78 rad.     (12.21)

Thus a complete mathematical expression for the vertical displacement of the motor

operating at a constant speed of 1000 rpm is

x\left(t\right)=0.00167\sin \left(104.7t-2.78\right)m .     (12.22)

Now, the force transmitted through the shock mounts is

F_{t}=b\dot{x}+kx=800\dot{x} +50,000x .     (12.23)

The transfer function relating the transmitted force to the input force F_{m} is

T_{F}\left(s\right) =\frac{F_{T}\left(s\right)}{T_{m}\left(s\right)} =\frac{800s + 50,000}{25s^{2} + 800s + 50,000} .      (12.24)

The frequency-response transfer function relating the two forces is

T_{F}\left(j\omega \right) = \frac{800 j\omega + 50,000}{−25\omega ^{2} + 800 j\omega + 50,000} ,      (12.25)

and for the frequency of vibration,

T_{F}\left(j\omega \right)\mid _{\omega=104.07 } = \frac{50,000 + j83,760}{−224,052 + j83,760} .      (12.26)

Hence, the magnitude of the force transfer function is

\mid T_{F}(j\omega)\mid \mid _{\omega=104.07 } =\frac{\sqrt{50,000^{2}+83,760^{2}}}{\sqrt{\left(−224,052\right) ^{2}+83,760^{2}}}=0.41 .     (12.27)

Thus the amplitude of the force transmitted through the shock mounts will be equal to just over 40 percent of the amplitude of the force produced by the imbalance of the rotating elements in the motor.