Newton's second law is an ODE — and dimensions tell you the answer's shape before you solve it

Idea. The classification is decided by which of \(t, v, x\) actually appears on the right-hand side. If more than one appears, none of the three moves works.

(a) \(m\ddot x = -mg - b\dot x\). The right-hand side depends on \(\dot x\) only: case \(F(v)\). First reduction: write \(v = \dot x\), giving the first-order equation \(m\dot v = -mg - bv\), and separate, \[ t \;=\; \int_{v_0}^{v}\frac{m\,\mathrm dw}{-mg - bw}, \] valid on any interval avoiding the zero \(w = -mg/b\).

(b) \(m\ddot x = -\kappa x^3\). Position only: case \(F(x)\). First reduction: multiply by \(\dot x\) and integrate, \(\tfrac12 m\dot x^2 + \tfrac14\kappa x^4 = E\).

(c) \(m\ddot x = F_0 e^{-t/\tau}\). Time only: case \(F(t)\). First reduction: integrate once, \(\dot x(t) = v_0 + \frac{F_0\tau}{m}\big(1 - e^{-t/\tau}\big)\).

(d) \(m\ddot x = -b\dot x - \kappa x\) depends on both \(x\) and \(\dot x\). None of the three. Multiplying by \(\dot x\) gives \(\frac{\mathrm d}{\mathrm dt}\big(\tfrac12 m\dot x^2 + \tfrac12\kappa x^2\big) = -b\dot x^2\), which is not zero, so there is no first integral to exploit; separating in \(v\) is impossible because \(x\) is still present.

(e) \(m\ddot x = -\kappa x(1 + \epsilon\cos\omega t)\) depends on both \(x\) and \(t\). None of the three, for the mirror-image reason: the equation is not invariant under time translation, so there is no energy integral.

What would have to be true. For (d): \(b = 0\), which restores case \(F(x)\); or \(\kappa = 0\), which gives case \(F(v)\). For (e): \(\epsilon = 0\). Both equations are nevertheless solvable — (d) by unit 06's linear theory, (e) not in closed form at all (it is Mathieu's equation, and unit 21 says what can be done with it). "Not reducible to quadrature" is not the same as "hopeless"; it is the reason units 01, 02 and 06 exist.

Check. Dimensions of (c)'s first reduction: \([F_0\tau/m] = \mathsf{MLT^{-2}}\cdot\mathsf T/\mathsf M = \mathsf{LT^{-1}}\), a velocity. ✓ Limit of (c) as \(\tau\to\infty\): \(1 - e^{-t/\tau} \to t/\tau\), so \(\dot x \to v_0 + F_0 t/m\), constant force. ✓

Idea. Write the matrix, find the rank by spotting which columns are obviously independent, solve \(Da = 0\).

(a) mass on a spring

\([\kappa] = \mathsf{MT^{-2}}\) because \(\kappa x\) is a force. With quantities \((T, m, \kappa, a)\): \[ D = \begin{array}{c|cccc} & T & m & \kappa & a\\\hline \mathsf M & 0&1&1&0\\ \mathsf L & 0&0&0&1\\ \mathsf T & 1&0&-2&0 \end{array} \] Columns \(m, a, T\) are \(e_1, e_2, e_3\), so \(r = 3\) and \(n - r = 4 - 3 = 1\). Solving \(Da = 0\): the \(\mathsf L\) row gives \(a_a = 0\); the \(\mathsf M\) row gives \(a_m = -a_\kappa\); the \(\mathsf T\) row gives \(a_T = 2a_\kappa\). Taking \(a_\kappa = 1\), \[ \pi = \frac{T^2\kappa}{m} \quad\Longrightarrow\quad T = C\sqrt{m/\kappa}. \] The amplitude drops out entirely — the oscillator is isochronous, and with one group the conclusion is complete up to the single number \(C\) (which is \(2\pi\)). The deck proves the same thing twice more: from mechanical similarity, since \(-\kappa x\) is homogeneous of degree 1 so \(T\propto a^{0}\) (Mechanical similarity: scaling the equation, not the units), and by evaluating the period integral (The period integral).

(b) deep-water wave

Quantities \((v, g, \lambda, \rho)\): \[ D = \begin{array}{c|cccc} & v & g & \lambda & \rho\\\hline \mathsf M & 0&0&0&1\\ \mathsf L & 1&1&1&-3\\ \mathsf T & -1&-2&0&0 \end{array} \] \(\rho\) is the only column with a nonzero \(\mathsf M\) entry, so it is independent of the others; \(\lambda\) and \(v\) are independent in the \((\mathsf L,\mathsf T)\) plane. Hence \(r = 3\) and \(n - r = 1\). The \(\mathsf M\) row forces \(a_\rho = 0\); the \(\mathsf T\) row gives \(a_v = -2a_g\); the \(\mathsf L\) row then gives \(a_\lambda = -a_v - a_g = a_g\). With \(a_g = 1\), \[ \pi = \frac{g\lambda}{v^2} \quad\Longrightarrow\quad v = C\sqrt{g\lambda}. \] Density drops out, and long waves travel faster than short ones — the reason a distant storm announces itself with swell before chop.

(c) escape speed

Quantities \((v, G, M, R)\), \([G] = \mathsf{M^{-1}L^3T^{-2}}\): \[ D = \begin{array}{c|cccc} & v & G & M & R\\\hline \mathsf M & 0&-1&1&0\\ \mathsf L & 1&3&0&1\\ \mathsf T & -1&-2&0&0 \end{array} \] Columns \(M, R, v\) give \(\mathsf M\), then \(\mathsf L\), then \(\mathsf T\) in turn, so \(r = 3\) and \(n - r = 1\). \(\mathsf M\): \(a_M = a_G\). \(\mathsf T\): \(a_v = -2a_G\). \(\mathsf L\): \(a_v + 3a_G + a_R = 0\), so \(a_R = -a_G\). With \(a_G = 1\), \[ \pi = \frac{GM}{v^2R} \quad\Longrightarrow\quad v = C\sqrt{GM/R}. \] (\(C = \sqrt2\); unit 08 supplies it.)

Check. Dimensions of each conclusion: \([\sqrt{m/\kappa}] = (\mathsf M/\mathsf{MT^{-2}})^{1/2} = \mathsf T\) ✓; \([\sqrt{g\lambda}] = (\mathsf{LT^{-2}}\cdot\mathsf L)^{1/2} = \mathsf{LT^{-1}}\) ✓; \([\sqrt{GM/R}] = (\mathsf{M^{-1}L^3T^{-2}}\cdot\mathsf M/\mathsf L)^{1/2} = \mathsf{LT^{-1}}\) ✓.

Idea. Case \(F(v)\): separate, invert, integrate once more. Careful with signs — \(v\) is positive upward and both gravity and drag act downward on the way up.

(a)

With \(x\) measured upward, \(m\dot v = -mg - bv\). Write \(\tau = m/b\) and \(v_\infty = mg/b = g\tau\) (the terminal speed for the downward motion). Then \(\dot v = -(v + v_\infty)/\tau\), and separating, \[ \int_{v_0}^{v}\frac{\mathrm dw}{w + v_\infty} = -\frac{t}{\tau} \;\Longrightarrow\; \ln\frac{v + v_\infty}{v_0 + v_\infty} = -\frac t\tau \;\Longrightarrow\; v(t) = -v_\infty + (v_0 + v_\infty)e^{-t/\tau}. \] The inversion is legitimate because \(w + v_\infty > 0\) throughout the upward leg, so the integrand never blows up. Setting \(v = 0\), \[ t_\uparrow = \tau\ln\!\Big(1 + \frac{v_0}{v_\infty}\Big),\qquad h = \int_0^{t_\uparrow}\! v\,\mathrm dt = v_\infty\tau\left[\frac{v_0}{v_\infty} - \ln\!\Big(1+\frac{v_0}{v_\infty}\Big)\right]. \] (The second follows from \(\int_0^t v = -v_\infty t + \tau(v_0+v_\infty) (1 - e^{-t/\tau})\) together with \(e^{-t_\uparrow/\tau} = v_\infty/(v_0+v_\infty)\).)

Numbers. \(\tau = m/b = 0.00500/0.0500 = 0.1000\) s, \(v_\infty = g\tau = 0.9810\ \mathrm{m\,s^{-1}}\), \(v_0/v_\infty = 2.0387\): \[ t_\uparrow = 0.1000\times\ln(3.0387) = \mathbf{0.1111\ s},\qquad h = 0.09810\times(2.0387 - 1.1113) = \mathbf{0.0910\ m}. \]

(b)

\(t_\uparrow = \tau\ln(1 + v_0/(g\tau))\) and \(\ln(1+z) < z\) for every \(z > 0\), so \(t_\uparrow < \tau\cdot v_0/(g\tau) = v_0/g\). Physically: on the way up, drag acts downward alongside gravity, so the deceleration is greater than \(g\) at every instant, so the bead stops sooner than it would in vacuum.

(c)

Put \(\epsilon = v_0/(g\tau) = bv_0/(mg)\). Then \(h = g\tau^2\big[\epsilon - \ln(1+\epsilon)\big]\) and \(\epsilon - \ln(1+\epsilon) = \tfrac12\epsilon^2 - \tfrac13\epsilon^3 + O(\epsilon^4)\), so \[ h = \frac{v_0^2}{2g} - \frac{b\,v_0^3}{3\,m\,g^2} + O(b^2). \] As \(b \to 0\), \(h \to v_0^2/2g\), the vacuum apex. ✓

Honest warning about the numbers. For these data \(\epsilon = 2.04\), which is not small: the vacuum apex is \(0.2039\) m, the first correction is \(0.2771\) m, and the "corrected" answer would be negative. The true \(h\) is \(0.0910\) m, less than half the vacuum value. The expansion is a statement about the limit \(b \to 0\) and says nothing useful here — a good illustration that an asymptotic correction is only a correction where the parameter is small. Unit 21 is about exactly this distinction.

(d)

Slower on the way down. The descent takes \(0.1754\) s against an ascent of \(0.1111\) s, a ratio of \(1.578\). (Solve \(v_\infty(t - \tau(1-e^{-t/\tau})) = h\) numerically for the descent time.)

Argument without the numbers. Energy. On the way up the bead does work against gravity and against drag, so it arrives at the top having lost \(\tfrac12 mv_0^2 - mgh\) to drag. On the way down it starts from rest and drag again removes energy, so it returns to the throw point with speed \(v_1 < v_0\). It therefore covers the same distance \(h\) with a smaller speed at every corresponding height — the downward journey is everywhere slower than the upward one at the same height — so it takes longer.

Check. Dimensions: \([\tau] = \mathsf M/(\mathsf{MT^{-1}}) = \mathsf T\), \([v_\infty\tau] = \mathsf L\). ✓ Limit \(b\to0\): \(t_\uparrow \to v_0/g\) and \(h \to v_0^2/2g\), both vacuum values. ✓

Idea. Case \(F(v)\), with a quadratic zero of the force at \(v_\infty\).

(a)

\(v_\infty = \sqrt{Mg/c}\), so \(c = Mg/v_\infty^2 = 80.0\times9.81/55.0^2 = \mathbf{0.2594\ kg\,m^{-1}}\). Separating, \[ t = \int_0^v\frac{M\,\mathrm dw}{Mg - cw^2} = \frac1g\int_0^v\frac{\mathrm dw}{1 - (w/v_\infty)^2} = \frac{v_\infty}{g}\operatorname{artanh}\frac{v}{v_\infty}, \] using \(M/(Mg - cw^2) = \big(g(1 - (w/v_\infty)^2)\big)^{-1}\) and \(\int_0^z(1-u^2)^{-1}\mathrm du = \operatorname{artanh} z\). This is valid for \(0 \le v < v_\infty\), where the integrand is finite; at \(v = v_\infty\) the integral diverges, which is the statement that terminal speed is never attained. Inverting (legitimate: \(\operatorname{artanh}\) is strictly increasing with nonvanishing derivative on \((-1,1)\)), \[ v(t) = v_\infty\tanh\frac{gt}{v_\infty},\qquad x(t) = \int_0^t v = \frac{v_\infty^2}{g}\Big[\ln\cosh\frac{gs}{v_\infty}\Big]_0^t = \frac{v_\infty^2}{g}\ln\cosh\frac{gt}{v_\infty}, \] since \(\tfrac{\mathrm d}{\mathrm du}\ln\cosh u = \tanh u\) and \(\ln\cosh 0 = 0\).

(b)

\(v = 0.99\,v_\infty\) needs \(gt/v_\infty = \operatorname{artanh}0.99 = 2.6467\), so with \(v_\infty/g = 5.6065\) s, \[ t_{99} = \mathbf{14.84\ s},\qquad x_{99} = \frac{55.0^2}{9.81}\ln\cosh(2.6467) = \mathbf{604\ m}. \] Drag-free at the same instant: \(v = gt_{99} = 145.6\ \mathrm{m\,s^{-1}}\), against her actual \(0.99v_\infty = 54.45\ \mathrm{m\,s^{-1}}\) — 2.67 times larger — and \(x = \tfrac12 gt_{99}^2 = 1080\) m against \(604\) m, 1.79 times larger.

(c)

\(\cosh u = \tfrac12(e^u + e^{-u}) = \tfrac12 e^u(1 + e^{-2u})\), so \(\ln\cosh u = u - \ln 2 + \ln(1 + e^{-2u}) = u - \ln2 + O(e^{-2u})\). Hence \[ x(t) = \frac{v_\infty^2}{g}\Big(\frac{gt}{v_\infty} - \ln2\Big) + O(e^{-2gt/v_\infty}) = v_\infty t - \frac{v_\infty^2\ln2}{g} + o(1), \] with the constant lag \(v_\infty^2\ln2/g = 55.0^2\times0.6931/9.81 = \mathbf{213.7\ m}\). At \(t_{99}\) the asymptotic formula gives \(55.0\times14.8385 - 213.7 = 602.4\) m against the exact \(603.9\) m: an error of \(1.5\) m, or 0.26 per cent, already.

Check. Dimensions: \([v_\infty^2/g] = \mathsf{L^2T^{-2}}/\mathsf{LT^{-2}} = \mathsf L\). ✓ Limit \(t \ll v_\infty/g\): \(\tanh u \approx u\) gives \(v \approx gt\) and \(\ln\cosh u \approx u^2/2\) gives \(x \approx \tfrac12gt^2\) — free fall. ✓ Limit \(c \to 0\): \(v_\infty \to \infty\) and the same expansion holds for all fixed \(t\). ✓

(a)

The dimension of a product is the sum of the dimensions: if \(q\) has dimension vector \(u\) and \(q'\) has \(u'\), then under \(\lambda\) the product scales by \(\prod_i\lambda_i^{u_i}\prod_i\lambda_i^{u_i'} = \prod_i\lambda_i^{(u+u')_i}\). By induction the monomial \(\prod_j q_j^{a_j}\) has dimension vector \(\sum_j a_j D_{\cdot j} = Da\). A quantity is dimensionless exactly when its dimension vector is \(0\), that is, exactly when \(Da = 0\).

For the group statement: let \(\mathcal G\) be the set of formal monomials \(\prod_j q_j^{a_j}\) with \(a \in \mathbb{Q}^n\), under multiplication. The map \(\Phi : (\mathbb{Q}^n, +) \to \mathcal G\), \(a \mapsto \prod q_j^{a_j}\), is a surjective homomorphism, and it is injective if the \(q_j\) are treated as independent formal symbols. Restricting to dimensionless monomials restricts the domain to \(\ker D\) by the first paragraph, so \(\Phi|_{\ker D}\) is a group isomorphism \((\ker D, +) \cong (\text{dimensionless monomials}, \times)\). In particular the number of independent groups is \(\dim\ker D = n - r\), by rank–nullity.

(b)

If \(r = n\) then \(\ker D = \{0\}\), so by (a) the only dimensionless monomial is \(\prod q_j^0 = 1\).

Now let \(\tilde D = [\,D_{\cdot Q}\ \ D\,]\) be the enlarged matrix and suppose \(\operatorname{rank}\tilde D = n\). Since \(D\) already has rank \(n\) and is a submatrix, the column \(D_{\cdot Q}\) must lie in the column space of \(D\): there is \(c \in \mathbb{Q}^n\) with \(D_{\cdot Q} = Dc\), and it is unique because the columns of \(D\) are independent. Then \(\pi := Q\prod_j q_j^{-c_j}\) is dimensionless, since its dimension vector is \(D_{\cdot Q} - Dc = 0\). By Buckingham applied to the list \((Q, q_1,\dots,q_n)\) — which has \(n+1\) quantities and rank \(n\), hence one group — the relation determining \(Q\) is \(F(\pi) = 0\); as \(Q\) is determined by the \(q_j\), that relation fixes \(\pi\) at a constant value \(C\). Hence \(Q = C\prod_j q_j^{c_j}\), with \(c\) unique as shown. \(\square\)

(c)

Let \(D' = [\,D\ \ D_{\cdot\,n+1}\,]\) and \(r' = \operatorname{rank} D'\). Appending a column changes the rank by 0 or 1, and by 1 exactly when \(D_{\cdot\,n+1}\) is not in the column space of \(D\). The number of groups is \((n+1) - r'\).

• If \(D_{\cdot\,n+1} \in \operatorname{col} D\): \(r' = r\), so the count goes from \(n - r\) to \(n + 1 - r\): one more group.
• Otherwise: \(r' = r + 1\), so the count goes from \(n - r\) to \(n + 1 - (r+1) = n - r\): unchanged.

(d)

Adding an irrelevant quantity whose dimensions are already spanned by the others adds a group, by (c). Buckingham then gives \(F(\pi_1,\dots,\pi_{n-r}, \pi_{n+1-r}) = 0\), an unknown function of one more variable — which is a strictly weaker statement, because a function of \(m+1\) variables is far less constrained than one of \(m\). (The physics is not wrong: the true law simply does not depend on the extra group.)

Omitting a relevant quantity is different in kind. The remaining relation is then not dimensionally homogeneous in the reduced list — the hypothesis of the theorem fails — and the "conclusion" is a false statement rather than a weak one. The diagnostic is usually visible: the groups you get are vacuous, or the predicted scaling contradicts an experiment. The deck's example is the pendulum with \(g\) omitted, where the only group is \(\theta_0\) and the theorem says nothing at all about \(T\).

Check. Count on the pendulum: \(n = 5\), \(r = 3\), \(n - r = 2\). ✓ Adding the bob's radius \(a\), whose dimension \(\mathsf L\) is already in the column space: \(n = 6\), \(r = 3\), \(3\) groups — one more, as (c) predicts. ✓

(a)

In \((\mathsf M, \mathsf L, \mathsf T)\): \[ D = \begin{array}{c|cccc} & R & E & \rho & t\\\hline \mathsf M & 0&1&1&0\\ \mathsf L & 1&2&-3&0\\ \mathsf T & 0&-2&0&1 \end{array} \] The last three columns have determinant \(\det\begin{pmatrix}1&1&0\\2&-3&0\\-2&0&1\end{pmatrix} = -5 \ne 0\), so \(r = 3\) and \(n - r = 4 - 3 = 1\). Solving \(Da = 0\): \(\mathsf M\) gives \(a_\rho = -a_E\); \(\mathsf T\) gives \(a_t = 2a_E\); \(\mathsf L\) gives \(a_R = -2a_E + 3a_\rho = -5a_E\). With \(a_E = 1\), \[ \pi = \frac{Et^2}{\rho R^5}. \] One group, so \(\pi\) is a constant, say \(C^{-5}\), and \(R(t) = C(Et^2/\rho)^{1/5}\).

(b)

If \(R = At^p\) then \(p = \ln(R_2/R_1)/\ln(t_2/t_1)\): \[ p = \frac{\ln(185.0/106.5)}{\ln(62.0/15.0)} = \frac{0.55221}{1.41908} = \mathbf{0.3891}, \] against the predicted \(0.4000\). The discrepancy is 2.7 per cent over a factor of 4 in time and 1.7 in radius — the power law is confirmed.

(c)

\(E = \rho R^5/(C^5t^2)\) with \(C^5 = 1.033^5 = 1.1762\).

Mean \(\mathbf{14.9\ kt}\). The two frames agree to 8 per cent, which is the real test: a single frame could be a coincidence, two frames a factor of four apart in time agreeing is the scaling law working.

(d)

\(E \propto R^5\), so a factor \(\mu\) on \(R\) is \(\mu^5\) on \(E\). To turn \(14.89\) kt into \(21\) kt needs \(\mu = (21/14.89)^{1/5} = \mathbf{1.0712}\): a 7.1 per cent under-read.

It is close to 1 precisely because of the fifth power — the estimate of \(E\) is extraordinarily sensitive to \(R\), so a small measurement error covers a large discrepancy in yield. That cuts both ways: the exponent \(2/5\) is robust and was confirmed in (b), while the prefactor was never going to be good, since reading a luminous fireball's edge off a 1945 photograph to better than a few per cent is not possible. The other contributors are the ambient density (Trinity was fired at 1470 m altitude, where \(\rho\) is nearer \(1.0\) than \(1.25\ \mathrm{kg\,m^{-3}}\)) and the energy that goes into radiation and into the ground rather than into the blast wave.

Check. Dimensions: \([Et^2/\rho] = \mathsf{ML^2T^{-2}}\cdot\mathsf{T^2}/ \mathsf{ML^{-3}} = \mathsf L^5\), so \((Et^2/\rho)^{1/5}\) is a length. ✓ Limit \(t\to0\): \(R\to0\). ✓ Number: Taylor's own published figure was about 17 kt, which is what this method gives with \(C = 1\); the honest conclusion is "within about 30 per cent of a classified number, from a magazine photograph".

(a)

\(V\) is continuous, even, strictly increasing on \([0,\infty)\) with \(V(0) = 0\) and \(V\to\infty\); so for any \(E > 0\) the set \(\{V \le E\}\) is the bounded interval \([-a, a]\) with \(a = (E/k)^{1/n}\), and the motion is bounded with turning points \(\pm a\). By the period-integral theorem, \[ T(E) = \sqrt{2m}\int_{-a}^{a}\frac{\mathrm dx}{\sqrt{E - k|x|^n}} = 2\sqrt{2m}\int_{0}^{a}\frac{\mathrm dx}{\sqrt{E - kx^n}}, \] using the evenness. Substitute \(x = a\,\sigma\), so \(kx^n = E\sigma^n\): \[ T = 2\sqrt{2m}\;\frac{a}{\sqrt E}\int_0^1\frac{\mathrm d\sigma}{\sqrt{1-\sigma^n}} . \] Now put \(u = \sigma^n\), \(\sigma = u^{1/n}\), \(\mathrm d\sigma = \tfrac1n u^{1/n - 1}\mathrm du\): \[ \int_0^1\frac{\mathrm d\sigma}{\sqrt{1-\sigma^n}} = \frac1n\int_0^1 u^{\frac1n - 1}(1-u)^{-\frac12}\,\mathrm du = \frac1n B\!\Big(\frac1n,\frac12\Big). \] With \(a/\sqrt E = k^{-1/n}E^{1/n - 1/2}\), \[ T(E) = \frac{4}{n}\,\frac{1}{\sqrt2}\,\sqrt m\;k^{-1/n}\,E^{\frac1n-\frac12}\, B\!\Big(\frac1n,\frac12\Big), \] as claimed: the prefactor is \(2\sqrt{2m}\cdot\frac1n = \frac{2\sqrt2}{n}\sqrt m\), and \(2\sqrt2 = 4/\sqrt2\).

The integral is improper at \(\sigma = 1\) and converges, since near \(\sigma = 1\) the integrand is \(\asymp (1-\sigma)^{-1/2}\); equivalently \(B(p,\tfrac12)\) converges for \(p > 0\).

(b)

\(E^{1/n - 1/2}\) is constant in \(E\) exactly when \(1/n = 1/2\), that is \(n = 2\) — *the harmonic oscillator is the unique isochronous power-law potential*. Then \(B(\tfrac12,\tfrac12) = \int_0^1\mathrm du/\sqrt{u(1-u)} = \pi\) (substitute \(u = \sin^2\varphi\), which turns it into \(2\int_0^{\pi/2}\mathrm d\varphi\)), and \[ T = \frac42\cdot\frac{1}{\sqrt2}\sqrt m\,k^{-1/2}\,\pi = \pi\sqrt{2m/k}. \] For the standard form \(V = \tfrac12\kappa x^2\) we have \(k = \kappa/2\), so \(T = \pi\sqrt{4m/\kappa} = 2\pi\sqrt{m/\kappa}\). ✓

(c)

The constant is \((4/n)(1/\sqrt2)B(1/n,1/2)\):

• \(n = 1\): \(4\cdot\tfrac1{\sqrt2}\cdot 2 = 4\sqrt2 = \mathbf{5.6569}\), and \(T = 5.6569\sqrt m\,k^{-1}E^{1/2}\). (This is the bouncing-ball potential; the period grows like \(\sqrt E\), i.e. like the drop height's square root.)
• \(n = 4\): \(1\cdot\tfrac1{\sqrt2}\cdot 5.2441 = \mathbf{3.7081}\), and \(T = 3.7081\sqrt m\,k^{-1/4}E^{-1/4}\): a quartic well runs faster the more energy it has.

(d)

\(V = k|x|^n\) gives \(F = -V' = -nk|x|^{n-1}\operatorname{sgn}x\), homogeneous of degree \(n - 1\). The deck's result with exponent \(k_{\text{hom}} = n-1\) gives \(T\propto a^{(1 - (n-1))/2} = a^{1 - n/2}\). Since \(E = ka^n\), \(a\propto E^{1/n}\), so \(T\propto E^{(1/n)(1 - n/2)} = E^{1/n - 1/2}\). ✓ The two agree, and the relation that makes them agree is \(a \propto E^{1/n}\).

Check. Dimensions: \([k] = \mathsf{ML^{2-n}T^{-2}}\) (so that \(k|x|^n\) is an energy), hence \([\sqrt m k^{-1/n}E^{1/n-1/2}] = \mathsf M^{1/2}\cdot \mathsf M^{-1/n}\mathsf L^{(n-2)/n}\mathsf T^{2/n}\cdot \mathsf M^{1/n - 1/2}\mathsf L^{2/n-1}\mathsf T^{1-2/n} = \mathsf T\). ✓ Limit/special case: \(n = 2\) reproduces \(2\pi\sqrt{m/\kappa}\). ✓ Number: \(4\sqrt2 = 5.6569\) and \(5.2441/\sqrt2 = 3.7081\). ✓

Idea. Taylor with remainder turns \(E - V\) into a power of \(x - a\); the comparison test does the rest. The exponent \(-1/2\) is integrable and \(-1\) is not, and everything hinges on which one you get.

Write \(I(\delta) := \int_{a-\delta}^{a}\mathrm dx\big/\sqrt{\tfrac2m(E-V(x))} = \sqrt{m/2}\int_{a-\delta}^{a}(E - V(x))^{-1/2}\mathrm dx\), and put \(s = a - x \in (0,\delta]\).

(a) \(V'(a) \neq 0\): convergent

Since \(V < E = V(a)\) to the left of \(a\), \(V\) is increasing there, so \(V'(a) \ge 0\); with \(V'(a)\ne0\) write \(V'(a) = c > 0\). Taylor with Lagrange remainder about \(a\): \[ E - V(a - s) = V(a) - V(a-s) = c\,s - \tfrac12 V''(\xi)\,s^2 \quad\text{for some }\xi\in(a-s,a). \] Let \(M = \sup_{[a-\delta_0,a]}|V''|\) and choose \(\delta \le \min(\delta_0, c/M)\). Then \(\tfrac12|V''(\xi)|s^2 \le \tfrac12 Ms\cdot s \le \tfrac12 cs\), so \[ \tfrac12 c\,s \;\le\; E - V(a-s) \;\le\; \tfrac32 c\,s . \] Hence \((E-V)^{-1/2} \le \sqrt{2/(cs)}\) and \[ I(\delta) \;\le\; \sqrt{\frac m2}\int_0^\delta\sqrt{\frac{2}{c\,s}}\,\mathrm ds = \sqrt{\frac mc}\int_0^\delta s^{-1/2}\mathrm ds = 2\sqrt{\frac{m\delta}{c}} \;<\;\infty, \] valid for \(\delta \le \min(\delta_0, |V'(a)|/\sup|V''|)\). That is the required bound, and the comparison test gives convergence.

(b) \(V'(a) = 0\), \(V''(a) < 0\): divergent

Write \(V''(a) = -2\kappa\) with \(\kappa > 0\). Taylor to third order (using \(V \in C^3\)): \(E - V(a-s) = \kappa s^2 + R(s)\) with \(|R(s)| \le M's^3\) near 0. Choose \(\delta\) with \(M'\delta \le \kappa\); then \(E - V(a-s) \le 2\kappa s^2\), so \[ I(\delta) \;\ge\; \sqrt{\frac m2}\int_0^\delta\frac{\mathrm ds}{\sqrt{2\kappa}\,s} = \frac12\sqrt{\frac m\kappa}\int_0^\delta\frac{\mathrm ds}{s} = \infty . \]

(c)

For the pendulum, \(V(\theta) = mgl(1-\cos\theta)\) and the turning point is \(\theta_0\) with \(E = V(\theta_0)\).

• \(0 < \theta_0 < \pi\): \(V'(\theta_0) = mgl\sin\theta_0 \ne 0\), so (a) applies and the quarter-period is finite; the period is four times it, hence finite.
• \(\theta_0 = \pi\): \(V'(\pi) = 0\) and \(V''(\pi) = -mgl < 0\), so (b) applies and the time is infinite. This is the separatrix.

(d) the rate

Near the top write \(s = a - x\) and \(V = E_c - \kappa s^2 + O(s^3)\). At energy \(E = E_c - \Delta\) the turning point is at \(s_t\) with \(\kappa s_t^2 = \Delta\), i.e. \(s_t = \sqrt{\Delta/\kappa}\), and \(E - V = \kappa s^2 - \Delta\) to leading order. Then \[ \int_{s_t}^{\delta}\frac{\sqrt{m/2}\ \mathrm ds}{\sqrt{\kappa s^2 - \Delta}} = \sqrt{\frac{m}{2\kappa}}\int_{s_t}^{\delta}\frac{\mathrm ds}{\sqrt{s^2 - s_t^2}} = \sqrt{\frac{m}{2\kappa}}\;\operatorname{arcosh}\frac{\delta}{s_t}. \] For large \(z\), \(\operatorname{arcosh} z = \ln(2z) + O(z^{-2})\), and \(\delta/s_t = \delta\sqrt{\kappa/\Delta}\), so the time is \[ \sqrt{\frac{m}{2\kappa}}\Big[\tfrac12\ln\tfrac1\Delta + \ln(2\delta\sqrt\kappa)\Big] \;\sim\; \tfrac12\sqrt{\frac{m}{2\kappa}}\,\ln\frac1\Delta . \]

Numerical confirmation on the pendulum. Here the coordinate is \(\theta\), the effective mass is \(ml^2\), and \(\kappa = \tfrac12|V''(\pi)| = \tfrac12 mgl\), so \(\sqrt{m_{\text{eff}}/2\kappa} = \sqrt{l/g}\). All four legs of the period approach a turning point, so \(T \sim 4\cdot\tfrac12\sqrt{l/g}\ln(1/\Delta) = 2\sqrt{l/g}\,\ln(1/\Delta)\). Also \(\Delta = 2mgl - mgl(1-\cos\theta_0) = 2mgl\cos^2(\theta_0/2)\), so each factor of 10 in \(\pi - \theta_0\) is a factor of 100 in \(\Delta\) and should add \(2\sqrt{l/g}\ln 100 = 2\times0.31928\times4.6052 = 2.941\) s. Observed: \(10.766 - 7.826 = 2.940\) s and \(13.707 - 10.766 = 2.941\) s. ✓

Check. Dimensions of the bound in (a): \([\sqrt{m\delta/c}]\) with \([c] = [V'] = \mathsf{MLT^{-2}}\) gives \((\mathsf M\cdot\mathsf L/\mathsf{MLT^{-2}})^{1/2} = \mathsf T\). ✓ Special case: for \(V = \tfrac12\kappa x^2\) with turning point \(a\), (a)'s bound with \(c = \kappa a\), \(M = \kappa\), \(\delta = a\) gives \(2\sqrt{m a/(\kappa a)} = 2\sqrt{m/\kappa}\) for a quarter period, against the true \(\tfrac12\pi\sqrt{m/\kappa} = 1.571\sqrt{m/\kappa}\): the bound holds and is within 30 per cent. ✓

(a)

\(V(x) = -\int_0^x F = -\int_0^x(-\kappa s + \alpha s^3)\mathrm ds = \tfrac12\kappa x^2 - \tfrac14\alpha x^4\), with \(V(0) = 0\). ✓

Equilibria: \(V'(x) = \kappa x - \alpha x^3 = x(\kappa - \alpha x^2) = 0\), so \(x = 0\) and \(x = \pm x_\star\) with \(x_\star = \sqrt{\kappa/\alpha} = \sqrt{8.00/50.0} = \mathbf{0.400\ m}\).

Test with \(V''(x) = \kappa - 3\alpha x^2\):

• \(V''(0) = \kappa = 8.00 > 0\): strict local minimum, stable.
• \(V''(\pm x_\star) = \kappa - 3\kappa = -2\kappa = -16.0 < 0\): strict local maxima, unstable.

(b)

\(E_c = V(\pm x_\star) = \tfrac12\kappa\frac{\kappa}{\alpha} - \tfrac14\alpha\frac{\kappa^2}{\alpha^2} = \frac{\kappa^2}{4\alpha} = \frac{8.00^2}{4\times50.0} = \mathbf{0.320\ J}.\)

For \(E < E_c\) the component of \(\{V \le E\}\) containing the origin is an interval strictly inside \((-x_\star, x_\star)\): bounded, periodic motion. For \(E > E_c\) there is no turning point on either side of the origin (\(V\to-\infty\) as \(|x|\to\infty\)), so the particle escapes and in fact accelerates away for ever. \(E = E_c\) is the separatrix, reaching \(x_\star\) in infinite time by P8.

(c)

Starting at the origin, \(E = \tfrac12 mv_0^2\). Oscillation for \(\tfrac12 mv_0^2 < E_c\), escape for \(>\). The threshold is \[ v_c = \sqrt{2E_c/m} = \sqrt{2\times0.320/0.200} = \mathbf{1.789\ m\,s^{-1}} . \]

(d)

Near \(x = 0\), \(V \approx \tfrac12\kappa x^2\), so \(m\ddot x \approx -\kappa x\) with angular frequency \(\omega = \sqrt{\kappa/m} = \sqrt{8.00/0.200} = 6.3246\ \mathrm{rad\,s^{-1}}\) and \[ T \approx 2\pi/\omega = \mathbf{0.9935\ s}. \] It is only an approximation because the quartic term was dropped: the true potential is flatter than the parabola (the \(-\tfrac14\alpha x^4\) term is negative), so the restoring force is weaker and the true period is longer, growing without bound as \(E\to E_c\). By P7's method the amplitude dependence enters at order \((a/x_\star)^2\).

Check. Dimensions: \([\kappa^2/\alpha] = (\mathsf{MT^{-2}})^2/ (\mathsf{ML^{-2}T^{-2}}) = \mathsf{ML^2T^{-2}}\), an energy. ✓ Limit \(\alpha\to0\): \(x_\star\to\infty\) and \(E_c\to\infty\) — the lid disappears and every motion is bounded, as it must be for a pure harmonic well. ✓

(a)

\(\theta_0 = 45^\circ = 0.7853982\) rad, \(2\pi\sqrt{l/g} = 2.0060667\) s.

The three-term error is \(2.086256 - 2.086140 = \mathbf{0.116\ ms}\). The next term is \(173\theta_0^6/737280 = 5.51\times10^{-5}\), worth \(0.110\) ms — which accounts for 95 per cent of the residual, as a convergent series with rapidly shrinking positive terms should.

(b)

\(T\) increases with amplitude, so a smaller amplitude means a shorter period, which means more ticks per day: the clock gains.

Exact ratios: \(T(6^\circ)/T_0 = 1.00068582\) and \(T(4^\circ)/T_0 = 1.00030470\). The fractional change is \(1.00030470/1.00068582 - 1 = -3.809\times10^{-4}\), so over a day \[ 3.809\times10^{-4}\times 86400\ \mathrm s = \mathbf{32.9\ s\ per\ day\ gained}. \] (Using the two-term series instead: \((6^2 - 4^2)(\pi/180)^2/16 = 3.808\times10^{-4}\) — the same to four figures, since both amplitudes are small.)

(c)

Leading term only: \(\theta_0^2/16 = 10^{-3}\) gives \(\theta_0 = \sqrt{0.016} = 0.126491\) rad \(= \mathbf{7.247^\circ}\).

How far to trust it: at the 1-per-cent level the same estimate gives \(0.400\) rad \(= 22.918^\circ\) against the exact \(22.814^\circ\), an overestimate of \(0.46\) per cent. The neglected term is \(O(\theta_0^4)\) relative to a leading term \(O(\theta_0^2)\), so the relative error of the estimate scales as \(\theta_0^2\); dropping from \(22.9^\circ\) to \(7.2^\circ\) is a factor \(10\) in \(\theta_0^2\), predicting a \(0.046\)-per-cent overestimate. The exact answer is \(7.2441^\circ\), so the estimate is high by \(0.046\) per cent. ✓ The prediction of the error was itself right to two figures.

Check. Dimensions: the whole problem is a ratio of times, dimensionless. ✓ Limit \(\theta_0\to0\): every row of the table tends to \(2.0061\) s. ✓ Sign: all series coefficients are positive, so every correction lengthens the period — consistent with \(|\sin\theta| \le |\theta|\) making the true restoring force weaker than the linear one. ✓

(a)

Let \(\tilde x(t) = \alpha x(t/\beta)\). Differentiating twice by the chain rule, \(\ddot{\tilde x}(t) = (\alpha/\beta^2)\,\ddot x(t/\beta)\). Since \(x\) solves the equation, \(\ddot x(t/\beta) = F(x(t/\beta))/m\), so \[ m\ddot{\tilde x}(t) = \frac{\alpha}{\beta^2}\,F\big(x(t/\beta)\big). \] On the other hand \(F(\tilde x(t)) = F(\alpha x(t/\beta)) = \alpha^k F(x(t/\beta))\) by homogeneity. The two agree for all \(t\) exactly when \(\alpha/\beta^2 = \alpha^k\), i.e. \(\beta^2 = \alpha^{1-k}\). \(\square\)

(b)

If \(x\) is a bounded periodic motion of amplitude \(a\) and period \(T\), then \(\tilde x\) has amplitude \(\alpha a\) and period \(\beta T\), since \(\tilde x(t + \beta T) = \alpha x(t/\beta + T) = \alpha x(t/\beta) = \tilde x(t)\). So amplitudes scale by \(\alpha\) and periods by \(\beta = \alpha^{(1-k)/2}\): \(T\propto a^{(1-k)/2}\).

• Harmonic oscillator, \(F = -\kappa x\), \(k = 1\): \(T\propto a^0\). ✓ (And \(T = 2\pi\sqrt{m/\kappa}\) confirms it.)
• P7's \(V = k|x|^n\), \(F\) homogeneous of degree \(n-1\): \(T\propto a^{1 - n/2}\), and with \(E\propto a^n\) this is \(T\propto E^{1/n - 1/2}\), exactly P7(a). ✓

(c)

Put \(\tilde x_i(t) = \alpha x_i(t/\beta)\), so that \(|\tilde x(t)| = \alpha|x(t/\beta)|\). Then \[ m\ddot{\tilde x}_i(t) = \frac{m\alpha}{\beta^2}\ddot x_i(t/\beta) = -\frac{\alpha}{\beta^2}\,\frac{GMm\,x_i(t/\beta)}{|x(t/\beta)|^3}, \] while the equation demands \[ -\frac{GMm\,\tilde x_i(t)}{|\tilde x(t)|^3} = -\frac{GMm\,\alpha x_i(t/\beta)}{\alpha^3|x(t/\beta)|^3} = -\frac{1}{\alpha^2}\cdot\frac{GMm\,x_i(t/\beta)}{|x(t/\beta)|^3}. \] These agree for all \(t\) exactly when \(\alpha/\beta^2 = \alpha^{-2}\), that is \[ \boxed{\ \beta^2 = \alpha^3\ } \] — which is the case \(k = -2\) of (a), as it should be. So scaling every distance in an orbit by \(\alpha\) gives another orbit with all times scaled by \(\alpha^{3/2}\): *the square of the period is proportional to the cube of the linear size*, Kepler's third law.

(d)

With 1 yr \(= 365.25\) d:

Mars agrees to three parts in \(10^5\); Jupiter is off by nine parts in \(10^4\), thirty times worse.

Which assumption fails. The equation \(m\ddot x_i = -GMm x_i/|x|^3\) treats the Sun as fixed, i.e. assumes the planet's mass is negligible. Jupiter's mass is \(9.546\times10^{-4}\) of the Sun's, against \(3.2\times10^{-7}\) for Mars. The correct statement (unit 08, cited, where the two-body problem is reduced) has \(GM\) replaced by \(G(M + m)\), giving a period smaller by \((1 + m/M)^{-1/2}\): \[ 11.87288\times(1 + 9.546\times10^{-4})^{-1/2} = 11.8672\ \mathrm{yr}, \] which removes half the discrepancy and leaves \(4.4\times10^{-4}\) — the residue being perturbations by Saturn and the precision of the quoted semi-major axis. The scaling argument itself is exact; what it was applied to was an approximate equation.

Another way

(a) can be read as Buckingham on the equation rather than on the answer. Write the homogeneous force as \(F(x) = -\kappa x^{k}\), so that \(\kappa\) has dimension \(\mathsf{MT^{-2}L^{1-k}}\); the quantities are \(T, a, m, \kappa\). The rank is 3, so there is one group, and solving \(Da = 0\) gives \(\pi = T^2\kappa a^{k-1}/m\), hence \(T = C\sqrt{m/\kappa}\,a^{(1-k)/2}\) — the same exponent, now with the mass dependence visible too. Similarity and Buckingham are the same invariance read two ways; use whichever is quicker.

Check. Dimensions of \(\beta^2 = \alpha^{1-k}\): both sides are pure numbers, as they must be, since \(\alpha\) and \(\beta\) are ratios. ✓ Special case \(k = 0\) (constant force, free fall): \(\beta^2 = \alpha\), so a fall four times as high takes twice as long — Galileo. ✓ Number: Mars predicted \(1.88079\) yr against \(1.88085\) yr observed. ✓