Question

An application demands that a sinusoidal pressure variation of 250 Hz be measured with no more than 2% dynamic error. In selecting a suitable pressure transducer from a vendor catalog, you note that a desirable line of transducers has a fixed natural frequency of 600 Hz but that you have a choice of transducer damping ratios of between 0.5 and 1.5 in increments of 0.05. Select a suitable transducer.

Answer 1

Answer:

Explanation:

Given that:

f = 250 Hz

$\delta$= 2%

$f_n$= 600 Hz

$\zeta$ = 0.5 to 1.5  increment by 0.05

$F = A sin (Xt)$

For 250 Hz = 250 cycle/sec

$X = 2 \pi t 250$

$X = 500 \pi t$

$X = Asin (500 \pi t)$

$\omega = 250 \\ \\ \omega_n = 600$

M = 0.98 , 1.02

$M_{(w)}} = \sqrt{[1-(\frac{w}{w_n})^2 + ( 2\zeta \frac{w}{w_n})^2}$

$\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2$

$\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}$

$\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{0.98^2}-(1-(\frac{250}{600})^2)^2}$

$\zeta = 0.7183$

At 0.7183 value of damping ratio the error value was 2% at 0.98 value of M

$\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2$

$\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}$

$\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{1.02^2}-(1-(\frac{250}{600})^2)^2}$

$\zeta = 0.6330$

At  0.6330 value of damping ratio the error value was 2% at 1.02 value of M.

Hence, the damping ratio $\zeta$ of the transducer must be placed between 0.6330 to 0.7183