Answer :
Answer:
Explanation:
Given that:
f = 250 Hz
[tex]\delta[/tex]= 2%
[tex]f_n[/tex]= 600 Hz
[tex]\zeta[/tex] = 0.5 to 1.5 increment by 0.05
[tex]F = A sin (Xt)[/tex]
For 250 Hz = 250 cycle/sec
[tex]X = 2 \pi t 250[/tex]
[tex]X = 500 \pi t[/tex]
[tex]X = Asin (500 \pi t)[/tex]
[tex]\omega = 250 \\ \\ \omega_n = 600[/tex]
M = 0.98 , 1.02
[tex]M_{(w)}} = \sqrt{[1-(\frac{w}{w_n})^2 + ( 2\zeta \frac{w}{w_n})^2}[/tex]
[tex]\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2[/tex]
[tex]\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}[/tex]
[tex]\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{0.98^2}-(1-(\frac{250}{600})^2)^2}[/tex]
[tex]\zeta = 0.7183[/tex]
At 0.7183 value of damping ratio the error value was 2% at 0.98 value of M
[tex]\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2[/tex]
[tex]\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}[/tex]
[tex]\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{1.02^2}-(1-(\frac{250}{600})^2)^2}[/tex]
[tex]\zeta = 0.6330[/tex]
At 0.6330 value of damping ratio the error value was 2% at 1.02 value of M.
Hence, the damping ratio [tex]\zeta[/tex] of the transducer must be placed between 0.6330 to 0.7183