Question

A single-degree-of-freedom oscillator is subjected to harmonic excitation F(t)=$F_{0}cos\left(\omega t\right)$ as shown in the figure.


The non-zero value of $\omega$, for which the amplitude of the force transmitted to the ground will be $F_0$ is

• A. $\sqrt{\frac{k}{2m}}$
• B. $\sqrt{\frac{2k}{m}}$
• C. $\sqrt{\frac{k}{m}}$
• D. $2\sqrt{\frac{k}{m}}$

Answer 1

The correct option is B $\sqrt{\frac{2k}{m}}$

Given,
$F_T=F_0$

Then transmissibility

$ϵ=\frac{F_T}{F_0}=\frac{F_0}{F_0}=1$


$\frac{\sqrt{1+{\left(\frac{2\xi \omega }{{\omega }_n}\right)}^2}}{\sqrt{{(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2)}^2+{\left(\frac{2\xi \omega }{{\omega }_n}\right)}^2}}=1$

$1+{\left(\frac{2\xi \omega }{{\omega }_n}\right)}^2={(1-{\left(\frac{\omega }{{\omega }_n}\right)}^2)}^2+{\left(\frac{2\xi \omega }{{\omega }_n}\right)}^2$

$1-{\left(\frac{\omega }{{\omega }_n}\right)}^2=\pm 1$

Taking (- ve) sign

$\Rightarrow 1-{\left(\frac{\omega }{{\omega }_n}\right)}^2=-1$

$\Rightarrow {\left(\frac{\omega }{{\omega }_n}\right)}^2=\sqrt{2}$

$\Rightarrow \frac{\omega }{{\omega }_n}=\sqrt{2}=\sqrt{2}{\omega }_n=\sqrt{\frac{2k}{m}}$