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

Step-by-step 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-((w)/(w_n))^2 + ( 2\zeta (w)/(w_n))^2}`

`(1)/(M) = [1-((w)/(w_n))^2]+(2 \zeta (w)/(w_n))^2`

`\zeta = (w_n)/(2w)\sqrt{(1)/(M^2)-(1-((w)/(w_n))^2)^2}`

`\zeta = (600)/(2(250))\sqrt{(1)/(0.98^2)-(1-((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

`(1)/(M) = [1-((w)/(w_n))^2]+(2 \zeta (w)/(w_n))^2`

`\zeta = (w_n)/(2w)\sqrt{(1)/(M^2)-(1-((w)/(w_n))^2)^2}`

`\zeta = (600)/(2(250))\sqrt{(1)/(1.02^2)-(1-((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