The figure shows the mean power input $\bar{P}$ as a function of driving frequency for a mass on a spring with damping. (Driving force $=$
$F_{0} \sin \omega t$, where $F_{0}$ is held constant and $\omega$ is varied.) The $Q$ is high enough so that the mean power input, which is maximum at $\omega_{0}$, falls to half-maximum at the frequencies $0.98 \omega_{0}$ and $1.02 \omega_{0}$.
(a) What is the numerical value of $Q$ ?
(b) If the driving force is removed, the energy decreases according to the equation
$$
\boldsymbol{E}=\boldsymbol{E}_{0} e^{-\mathbf{q}}
$$
What is the value of $\gamma$ ?
(c) If the driving force is removed, what fraction of the energy is lost per cycle?
A new system is made in which the spring constant is doubled, but the mass and viscous medium are unchanged, and the same driving force $F_{0}$ sin $\omega t$ is applied. In terms of the corresponding quantities for the original system, find the values of the following:
(d) The new resonant frequency $\omega_{0}^{\prime}$.
(e) The new quality factor $Q^{\prime}$.
(f) The maximum mean power input $\bar{P}_{m}^{\prime} .$
(g) The total energy of the system at resonance, $E_{0^{\prime}}$.