Question

A mass of $50\text{kg}$ is suspended from a spring of stiffness $10\text{kN/m}$. It is set into oscillations and it is observed that two successive oscillations have amplitudes $10\text{mm}$ and $1\text{mm}$. Determine the damping ratio.

• A. $0.232$
• B. $0.344$
• C. $0.392$
• D. $0.446$

Answer 1

The correct option is B $0.344$

Let us consider two oscillations , one occuring ${}^{\prime }{m}^{\prime }$ cycles after the first oscillation.
Amplitude of first oscillation at time $t$ is given by $x_1=A{e}^{-\delta {\omega }_{0}t}.....\left(1\right)$
where $\delta =\frac{\gamma }{{\omega }_0}$
Amplitude of second oscillation after $m$ cycles (time $mT$) is given by $x_2=A{e}^{-\delta {\omega }_0(t+mT)}......\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we can write that,
$\frac{x_1}{x_2}=e^{\delta {\omega }_{0}mT}$
$\Rightarrow \delta {\omega }_{0}mT=\ln \left(\frac{x_1}{x_2}\right)......\left(3\right)$
Here, $\delta$ is known as damping ratio.
For successive amplitudes, $m=1$
$\therefore \delta {\omega }_{0}T=\ln \left(\frac{10}{1}\right)=2.3$
$\Rightarrow \delta {\omega }_0\left(\frac{2\pi }{{\omega }^{\prime }}\right)=2.3....\left(4\right)$
But we know that, ${\omega }^{\prime }=\sqrt{{\omega }_0^2-{\gamma }^2}$
From this we get, ${\omega }^{\prime }={\omega }_0\sqrt{1-{\delta }^2}$
where $\delta =\frac{\gamma }{{\omega }_0}$
From $\left(4\right)$ we can write that,
$\frac{2\pi \delta }{\sqrt{1-{\delta }^2}}=2.3$
Squaring on both sides ,
$\frac{39.4{\delta }^2}{(1-{\delta }^2)}=5.29$
$\Rightarrow {\delta }^2=0.118\Rightarrow \delta =0.344$
Thus, option (b) is the correct answer.