Mass — Question 9

Question 9

Measuring mass without a gravitational weighing signal

An ideal inertial comparator uses small horizontal oscillations. Its squared period is modeled by $$T^2=K(m+m_s),$$ where $m$ is the added sample mass, $m_s$ is the unknown effective carrier mass, and $K$ is an unknown positive constant. Take this relation as supplied; no oscillation theory is required.

Calibration gives $T=0.400\ \mathrm{s}$ with no sample and $T=0.600\ \mathrm{s}$ with a $0.400\ \mathrm{kg}$ reference sample. An unknown sample gives $T=0.700\ \mathrm{s}$. Treat all data as exact within the model, with $K$ and $m_s$ unchanged between runs.

1. Find $K$, its units, and $m_s$ from the two calibration runs.

2. Find the unknown mass. Derive an expression that eliminates both $K$ and $m_s$ in terms of the empty, reference, and unknown squared periods.

3. A student uses $m=m_{\mathrm{ref}}(T/T_{\mathrm{ref}})^2$. Find the resulting estimate and identify the contribution it ignores.

4. Would the supplied model permit mass measurement in a region with no gravitational weighing signal? What must be verified before transferring this calibration to that environment?

Question 9 - Solution

Measuring mass without a gravitational weighing signal

(a) Subtract squared periods to separate the carrier.

The empty run gives $0.400^2=Km_s$; the reference gives $0.600^2=K(0.400+m_s)$ in SI units. Subtracting yields $$K=\frac{0.360-0.160}{0.400}=\boxed{0.500\ \mathrm{s^2\,kg^{-1}}},\qquad m_s=\frac{0.160}{0.500}=\boxed{0.320\ \mathrm{kg}}.$$

(b) Remove the empty-device contribution before taking a ratio.

For empty, reference, and unknown periods $T_0,T_r,T_u$, $$T_u^2-T_0^2=Km_u,\qquad T_r^2-T_0^2=Km_r.$$ Thus $$\boxed{m_u=m_r\frac{T_u^2-T_0^2}{T_r^2-T_0^2}} =0.400\frac{0.490-0.160}{0.360-0.160}\ \mathrm{kg} =\boxed{0.660\ \mathrm{kg}}.$$

(c) Raw period ratios include the carrier in both runs.

$$\frac{T_u^2}{T_r^2}=\frac{m_u+m_s}{m_r+m_s},\qquad m_{\mathrm{naive}}=0.400\frac{0.490}{0.360}\ \mathrm{kg}\approx0.54444\ \mathrm{kg}.$$

The proposed expression gives $0.400(0.700/0.600)^2\approx\boxed{0.54444\ \mathrm{kg}}$, about 0.11556 kg too low. The correct raw ratio is $T_u^2/T_r^2=(m_u+m_s)/(m_r+m_s)$, not $m_u/m_r$. Neglecting the nonzero carrier mass therefore biases this inference.

(d) The model uses inertia rather than gravitational force.

No factor $g$ appears in the calibrated relation, so it permits a mass measurement without gravitational weighing. The actual apparatus must still retain the same elastic response, effective carrier mass, constraint geometry, and operating regime. Calibration runs with known masses in the new conditions would test that transfer. Absence of $g$ in an assumed model alone is not proof of an unchanged real instrument.