= Solution
For the supplied linear meter model let $\omega_0^2=C/I$ and $2\gamma=\Phi_0^2/(RI)$. If the applied step is $i_s$, the settled deflection is $\theta_\infty=\Phi_0i_s/C$. An <underdamped> response overshoots and rings about that value; a critically damped response reaches it monotonically without ringing; an <overdamped> response is also monotone from these initial conditions but has a slower final approach. The original plot uses the common final value and natural-frequency time scale.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-meter-response.png]
{title=Original normalized meter step responses and the critically damped amplitude and phase response}
The <critical step response of a galvanometer> has $\gamma=\omega_0$ and repeated homogeneous root $-\omega_0$. Apply $\theta(0)=\dot\theta(0)=0$ to the general constant-force solution to obtain
$$
\boxed{\theta(t)=\theta_\infty[1-(1+\omega_0t)e^{-\omega_0t}],\qquad t\ge0.}
$$
Its <derivative> is $\theta_\infty\omega_0^2t e^{-\omega_0t}$, with the sign of the step. It is the fastest nonoscillatory response within this second-order family at fixed $\omega_0$, avoiding a needle that repeatedly crosses the reading or an unnecessarily slow <overdamped> settling.
For sinusoidal drive choose $e^{i\omega t}$ <phasors>. Writing $F_0=\Phi_0i_0/I$, the response <amplitude> is
$$
\Theta=\frac{F_0}{\omega_0^2-\omega^2+2i\gamma\omega},\qquad\theta(t)=|\Theta|\cos(\omega t+\phi),
$$
$$
|\Theta|=\frac{F_0}{\sqrt{(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2}},\qquad \boxed{\phi=-\operatorname{atan2}(2\gamma\omega,\omega_0^2-\omega^2).}
$$
For positive $F_0$, choose the continuous lag branch from zero to $-\pi$ as <frequency> increases; an ordinary single-argument arctangent without the quadrant would give the wrong high-frequency <phase>.
For the <critical galvanometer transfer function>, the denominator factors as $(\omega_0+i\omega)^2$. Thus
$$
\boxed{\theta(t)=\frac{F_0}{\omega_0^2+\omega^2}\cos(\omega t+\phi),\qquad \phi=-2\arctan(\omega/\omega_0).}
$$
For $\omega\ll\omega_0$, the meter follows the quasi-static deflection $\Phi_0i_0/C$ with a small lag $\phi\simeq-2\omega/\omega_0$. For $\omega\gg\omega_0$, its <amplitude> is $F_0/\omega^2$ and its lag approaches $-\pi$.
The stated components give $\gamma=12.5$ s⁻¹ and, if critical, $\omega_0=12.5$ rad/s, or $f_0=1.99$ Hz. At 50 Hz, $\omega=314.16$ rad/s and
$$
\boxed{\frac{|\Theta(50\,\mathrm{Hz})|}{|\Theta(0)|}=\frac{12.5^2}{12.5^2+314.16^2}=1.58\times10^{-3},\qquad\phi=-175.44^\circ.}
$$
The needle barely responds compared with its DC sensitivity and its mean deflection for a pure sinusoid is zero in this linear model. \b[It is unsuitable for directly measuring 50-Hz <amplitude> or RMS current without an additional <rectifier> or another readout method.] The calculation treats $\Phi_0$ as the stipulated <torque>/back-emf coefficient of the given equation; it does not infer an unprovided winding geometry.
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