Problem

A block having a mass of 0.8 kg is suspended from a spring having a stiffness of 120 N/m....

A block having a mass of 0.8 kg is suspended from a spring having a stiffness of 120 N/m. If a dashpot provides a damping force of 2.5 N when the speed of the block is 0.2 m/s, determine the period of free vibration.

Step-by-Step Solution

Solution 1

Mass of block, \(m=0.8 \mathrm{~kg}\)

Stiffness of spring, \(k=120 \mathrm{~N} / \mathrm{m}\)

Damping force, \(F=2.5 \mathrm{~N}\)

The velocity of the block \(v=0.2 \mathrm{~m} / \mathrm{s}\)

Damping force \(F\) is proportional to the velocity of block by damping coefficient (C) times

$$ \begin{aligned} &F=C v \\ &F=C x \\ &2.5=C \times 0.2 \\ &C=12.5 \mathrm{~N} . \mathrm{s} / \mathrm{m} \end{aligned} $$

Let’s assume the system is displaced ‘y’ distance from equilibrium.

At equilibrium,

The forces acting along \(y\)-direction

$$ \begin{aligned} \sum F &=m \ddot{y} \\ m g-k\left(s_{0}\right) &=0 \\ s_{0} &=\frac{m g}{k} \end{aligned} $$

At distance \(y\), equation of motion is given by

$$ \begin{aligned} &m \ddot{y}=m g-k\left(s_{0}+y\right)-C \dot{y} \\ &m \ddot{y}+C \dot{y}+k y=0 \end{aligned} $$

Critical damping,

$$ \begin{aligned} &C_{C}=2 m \omega_{n} \\ &C_{C}=2 \times 0.8 \times \sqrt{\frac{120}{0.8}} \\ &C_{C}=19.5896 \mathrm{~N} . \mathrm{s} / \mathrm{m} \end{aligned} $$

Where \(C, system is under damped

$$ \begin{aligned} &\omega_{d}=\omega_{n} \sqrt{1-\left(\frac{C}{C_{C}}\right)^{2}} \\ &\omega_{d}=\sqrt{\frac{120}{0.8}} \sqrt{1-\left(\frac{12.5}{19.596}\right)^{2}} \\ &\omega_{d}=9.432 \mathrm{rad} / \mathrm{s} \end{aligned} $$

Period of free vibration.

$$ \begin{aligned} &T=\frac{2 \pi}{\omega_{d}} \\ &T=\frac{2 \pi}{9.432} \\ &T=0.666 \mathrm{sec} \end{aligned} $$

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