The phase step between adjacent slits isFor equal amplitudes $a_0$, the six unmodified contributions form a closed regular hexagon in the complex plane:Reversing one slit at an array minimum replaces slit $j$'s contribution by its negative, so $S=S_0-2a_0{e}^{ij\pi /3}$. ConsequentlyThis is independent of which slit is reversed. The result uses the far-field, narrow-slit and small-angle approximations; retaining $\sin \theta$ changes the phase step only by about $7\times 10^{-7}$ radians.

To find the magnitude uncertainty of a series RL impedance, put $X=\omega L$. Then $z=|Z|=\sqrt{R^2+X^2}=168.003\Omega$ andIf the quoted errors are independent standard uncertainties, ${\sigma }_R=1\Omega$ and ${\sigma }_X=6\Omega$ giveIf they denote strict bounds instead, the first-order bound is $(R\Delta R+X\Delta X)/z=5.42\Omega$. The source does not specify their statistical interpretation, so these two error conventions should not be conflated.

For a mass change in a viscously damped spring oscillator, use linear viscous damping $m\ddot{x}+b\dot{x}+kx=mg$, holding the spring constant $k$ and damping coefficient $b$ fixed. Measuring $x$ from $mg/k$ removes the constant force. The envelope is $e^{-\gamma t}$, whereDoubling $m$ halves both $k/m$ and $\gamma$. ThereforeNeglecting damping in the frequency gives $0.5/\sqrt{2}=0.35355\mathrm{Hz}$, indistinguishable at the supplied precision. The release condition sets the phase and new initial amplitude, not these envelope ratios. A different mass-dependent drag law would require its damping coefficient to be specified.

For drum-frequency dimensional scaling, let $f\propto r^p{\sigma }^q{T}^s$, where $\sigma$ is surface mass density. The dimensions are $[\sigma ]=M{L}^{-2}$ and $[T]=M{T}^{-2}$, so matching powers gives $s=1/2$, $q=-1/2$ and $p=-1$:The dimensionless normal mode constant $C$ is unchanged for the same circular shape and boundary condition. The doubled tension does not offset the doubled radius.

The shell theorem gives the gravity of a concentric hollow sphere as $\mathbf{g}(r)=-GM(<r)\hat{\mathbf{r}}/r^2$. There is no enclosed mass at the cavity boundary, whereas the entire shell has mass $M=\frac{4\pi \rho }{3}[(2a{)}^3-a^3]=\frac{28\pi }{3}\rho a^3$. ThusBoth nonzero fields point towards the centre. The field is continuous at each boundary; there is no surface mass sheet.