Physics - EMI/AC Question with Solution | TestHub

P: The displacement amplitude for forced oscillations is $A=\frac{F_0}{\sqrt{m^2({\omega }_0^2-{\omega }^2{)}^2+b^2{\omega }^2}}$. When $\omega ={\omega }_0$, $A=\frac{F_0}{b{\omega }_0}$. Thus (P)-(1). Q: The phase lag $\varphi$ is given by $\tan \varphi =\frac{b\omega }{m({\omega }_0^2-{\omega }^2)}$. For $\omega \gg {\omega }_0$, $\tan \varphi \approx \frac{b\omega }{-m{\omega }^2}=-\frac{b}{m\omega }$. As $\omega \to \infty$, $\tan \varphi \to 0$. Since the denominator $({\omega }_0^2-{\omega }^2)$ is negative, $\varphi$ approaches $\pi$. Thus (Q)-(5). R: The average power dissipated by the dashpot is $P_{avg}=\frac{1}{2}b{A}^2{\omega }^2$. At displacement resonance, $A=\frac{F_0}{b{\omega }_R}$, where ${\omega }_R=\sqrt{{\omega }_0^2-(b/2m{)}^2}$. Substituting this, $P_{avg}=\frac{1}{2}b{\left(\frac{F_0}{b{\omega }_R}\right)}^2{\omega }_R^2=\frac{F_0^2}{2b}$. Thus (R)-(3). S: The velocity amplitude is maximum at the natural frequency ${\omega }_0$. This is known as velocity resonance. Thus (S)-(6).