A 25 kg mass is resting on a spring of 5 kN/m stiffness and a dashpot of 150 N.s/m damping coefficient in parallel. If a velocity of 0.1 m/s is applied to the mass at the rest position, what will be its displacement from the equilibrium position at the end of first second?

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Accepted Answer

[latex] ext { Given : } m =25 kg , k =5 kN / M , c =150 N \cdot s / m , \dot{x}( o )=0.1 m / s [/latex].

Undamped natural frequency, [latex] \omega_{n}=\sqrt{\frac{k}{m}} [/latex].

[latex] =\sqrt{\frac{5000}{25}}=14.14 rad / s [/latex].

Critical damping coefficient, [latex] c_{c}=2 m \omega_{n} [/latex].

[latex] =2 imes 25 imes 14.14=707 N \cdot s / m [/latex].

Damping factor, [latex] \zeta=\frac{c}{c_{c}}=\frac{150}{707}=0.212 [/latex].

Damped natural frequency, [latex] \omega_{d}=\omega_{n} \sqrt{1-\zeta^{2}} [/latex].

[latex] =14.14 \sqrt{1-(0.212)^{2}} [/latex].

[latex] =13.82 rad / s [/latex].

[latex] x(t)=e^{-\zeta \omega_{n} t}\left[A \sin \omega_{d} t+B \cos \omega_{d} t ight] [/latex].

[latex] ext { Now } x(0)=0 ext {. Hence } B=0 [/latex].

[latex] \dot{x}(t)=e^{-\zeta \omega_{n} t}\left[A \omega_{d} \cos \omega_{d} t ight]-\zeta \omega_{n} e^{-\zeta \omega_{n} t} A \sin \omega_{d} t [/latex].

[latex] 0.1=A \omega_{d} [/latex].

[latex] A=\frac{0.1}{13.82}=0.007236 m [/latex].

[latex] x(t)=0.007236 \cdot e^{-0.212 imes 14.14 t} \cdot \sin 13.82 t [/latex].

[latex] x(1)=0.343 mm [/latex].

Question 17.9: A 25 kg mass is resting on a spring of 5 kN/m stiffness and ...

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