Physics - EMI/AC Question with Solution | TestHub

P: The logarithmic decrement $\delta$ is defined as $\ln (A_n/A_{n+1})$, where $A_n$ is the amplitude of the $n$ -th oscillation. For an underdamped oscillator, $\delta =\frac{b{T}^{\prime }}{2m}$, where $T^{\prime }$ is the period of damped oscillation. Thus (P)-(1). Q: The energy of a freely decaying underdamped oscillator decays as $E(t)=E_0{e}^{-\frac{b}{m}t}$. After $N$ periods, $t=N{T}^{\prime }$, so the ratio is $E(N{T}^{\prime })/E_0=e^{-N\frac{b{T}^{\prime }}{m}}$. Thus (Q)-(3). R: Instantaneous power supplied by the driving force is $P=F\cdot v$. Given $F=F_0\sin (\omega t)$ and $v=A\omega \cos (\omega t-\varphi )$, $P=F_0\sin (\omega t)A\omega \cos (\omega t-\varphi )$. Thus (R)-(5). S: At resonance, the displacement lags the driving force by $\frac{\pi }{2}$. If the driving force is $F_0\sin (\omega t)$, then the displacement is $x=A\sin (\omega t-\frac{\pi }{2})=-A\cos (\omega t)$. The velocity is $v=\frac{dx}{dt}=A\omega \sin (\omega t)$. Since the driving force is $F_0\sin (\omega t)$, the velocity is in phase with the driving force, so the phase difference is $0$. Thus (S)-(4).