An Angular Simple Harmonic Oscillator — Question 9

Question 9

A bias torque and a torque gradient leave different signatures

A torsion balance of known inertia $I$ and elastic constant $\kappa$ experiences an additional torque $\tau_e=\tau_0+c\theta$ near its operating point. The total torque is $\tau_0-(\kappa-c)\theta$, with $\kappa>c$.

1. Find the equilibrium angle and small-oscillation frequency.

2. A test doubles the equilibrium angle while leaving the period unchanged. Under this linear model with fixed $I$ and $\kappa$, which parameter changed? What additional measurement would a changed period provide?

Question 9 - Solution

A bias torque and a torque gradient leave different signatures

(a) Derivation and calculation

$$I\ddot\theta=\tau_0+c\theta-\kappa\theta=\tau_0-(\kappa-c)\theta.$$$$\boxed{\theta_e=\frac{\tau_0}{\kappa-c}},\qquad\boxed{\Omega^2=\frac{\kappa-c}{I}},\quad\kappa>c.$$$$q=\theta-\theta_e\Rightarrow I\ddot q+(\kappa-c)q=0.$$

(b) Result and interpretation

For fixed $I,\kappa$, an unchanged period implies unchanged $c$:$$T=2\pi\sqrt{I/(\kappa-c)},\quad\theta_e^{\prime}=2\theta_e\Rightarrow\boxed{\tau_0^{\prime}=2\tau_0}.$$$$\boxed{c=\kappa-\frac{4\pi^2I}{T^2}},\qquad\boxed{\tau_0=\frac{4\pi^2I}{T^2}\theta_e}.$$

Physical check

The equilibrium measures the torque offset relative to stiffness; the period independently determines the torque slope.