A mass of 5 kg hangs from a spring and makes damped oscillations. If the time of 50 complete oscillations is found to be 20 s, and the ratio of the first downward displacement to the sixth is found to be 22.5, find the stiffness of the spring and the damping coefficient.

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Accepted Answer

Given: [latex] m=5 kg , f_{d}=\frac{50}{20}=2.5 Hz , \frac{x_{1}}{x_{6}}=22.5 [/latex].

Logarithmic decrement, [latex] \delta=\left(\frac{1}{n} ight) \ln \left(\frac{x_{o}}{x_{n}} ight)=\left(\frac{1}{5} ight) \ln \left(\frac{x_{1}}{x_{6}} ight)=\left(\frac{1}{5} ight) \ln 22.5=0.6227 [/latex].

[latex] \delta=\frac{2 \pi \zeta}{\left(1-\zeta^{2} ight)^{1 / 2}} [/latex].

[latex] 0.6227=\frac{2 \pi imes \zeta}{\left(1-\zeta^{2} ight)^{1 / 2}} [/latex].

[latex] 1-\zeta^{2}=101.8 \zeta^{2} [/latex].

[latex] \zeta=0.0986 [/latex].

[latex] \omega_{d}=2 \pi f_{d}=2 \pi imes 2.5=15.708 rad / s [/latex].

[latex] \omega_{n}=\frac{\omega_{d}}{\left(1-\zeta^{2} ight)^{1 / 2}}=\frac{15.708}{\left[1-(0.0986)^{2} ight]^{1 / 2}}=15.785 rad / s [/latex].

Stiffness of spring, [latex] k=m \omega_{n}^{2}=5 imes(15.785)^{2}=1245.83 N / m [/latex].

Critical damping coefficient, [latex] c_{c}=2 m \omega_{n}=2 imes 5 imes 15.785=157.85 N \cdot s / m [/latex].

Damping coefficient, [latex] c=\zeta c_{c}=0.0986 imes 157.85=15.564 N \cdot s / m [/latex].

Question 17.6: A mass of 5 kg hangs from a spring and makes damped oscillat...

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