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Consider a simple mass-spring-friction system as given in the figure
k1 = spring constant f = friction
M = mass F = force
x = displacementThe transfer function X(s) of the given system will be F(s)
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1 Ms2 + fs + K1.K2 -
1 Ms2 + fs + K1 + K2 -
1 Ms2 + fs + K1.K2 K1 + K2 -
K2 Ms2 + fs + K1
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Correct Option: B
| F = M. | + f | + (K1 + K2)x | ||
| dt2 | dt |
∴ F(s) = Ms2 X(s) + f. s X(s) + (K1 + K2) X(s)
| ⇒ | = | |||
| F(s) | Ms2 + fs + (K1 + K2) |
