A mass weighing 13 lb stretches a spring 4.5 in. The mass is also attached to a damper with coefficient Y. Determine the value of Y for whic

Question

A mass weighing 13 lb stretches a spring 4.5 in. The mass is also attached to a damper with coefficient Y. Determine the value of Y for which the system is critically damped. Assume that g = 32 ft/s^2.

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Arya 2 weeks 2021-09-09T09:08:21+00:00 1 Answer 0

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    2021-09-09T09:09:46+00:00

    Answer:55.227

    Step-by-step explanation:

    Given data

    mass \left ( m\right )=13lb

    spring stretches by \left ( x\right )=4.5in=0.375 ft

    g=32ft/s^2

    Now Spring constant K

    mg=kx

    k=\frac{13\times 32}{y}=58.667lb/ft

    For critical damping \zeta =1

    2\times \zeta \times \omega_n =\frac{c}{m}

    and \omega_n=\sqrt {\frac{k}{m}}

    \omega _n=2.1241rad/s

    substituting values

    2\times 1 \times 2.124 =\frac{c}{m}

    c=55.227 lb.sec/ft

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