Hooke's law

Why a jump needs a model

• A bungee rope stores energy as it stretches. Its pull changes during the jump, so constant-acceleration formulae are usually unsuitable.
• Hooke's law 胡克定律 models force proportional to extension within a stated range. Real ropes need not follow it over a whole jump. Here use ideal light elastic 弹性 strings and springs with no energy loss unless stated.

Measure extension from natural length

• The natural length 自然长度 $L$ is the unloaded length. If the actual length is $\ell$, the extension 伸长量 is $x=\ell-L$, not $\ell$. A taut elastic string has tension $T=kx$.
• With modulus of elasticity 弹性模量 $\lambda$, write $k=\lambda/L$, so $T=\lambda x/L$. For a spring the signed restoring force opposes displacement from its natural length.

Modulus and stiffness have different units

• Stiffness 劲度系数 $k$ is measured in $\dfrac{\text{N}}{\text{m}}$; this string model's $\lambda$ is measured in newtons. It includes material and cross-section effects; it is not Young modulus, whose unit is pascals.
• Cut a uniform string into two equal pieces: each has the same $\lambda$ but half the natural length, so each piece has twice the stiffness. Do not keep $k$ fixed when changing $L$ in this way.

A string and a spring have different branches

• A string pulls only when taut 绷紧. For $\ell\leq L$ it is slack 松弛: $T=0$ and its stored elastic energy is zero. A negative value of $\ell-L$ is not a negative string tension.
• An ideal spring can pull in extension and push in compression 压缩. Its signed $x$ may be negative, while the force magnitude is $k|x|$. Continue to use the spring law only within the stated model range.

Store the area under the graph

• Stretching slowly from natural length does work equal to the triangular area under the force–extension line. The stored elastic potential energy 弹性势能 is

• For a spring this also applies in compression. For a string use it only when $x\geq0$; the slack branch has zero energy. Doubling extension doubles tension but multiplies energy by four. A proof of the energy formula is not required.

Use force and energy on the same string

• A string has $L=2\ \text{m}$, $\lambda=50\ \text{N}$ and extension $0.5\ \text{m}$. Its actual length is $2.5\ \text{m}$. First use $T=\lambda x/L$, giving $T=50(0.5)/2=12.5\ \text{N}$.
• For stored energy use $E_e=\lambda x^2/(2L)$, giving $E_e=50(0.5)^2/4=3.125\ \text{J}$. Force and energy have different units; do not substitute actual length for extension.

Infer a natural length from measurements

• A string has tension $6\ \text{N}$ at length $1.3\ \text{m}$ and $10\ \text{N}$ at length $1.5\ \text{m}$. Since $T=k(\ell-L)$, the gradient is $k=(10-6)/(1.5-1.3)=20\ \dfrac{\text{N}}{\text{m}}$.
• Then $L=\ell-T/k=1.3-6/20=1\ \text{m}$ and $\lambda=kL=20\ \text{N}$. Check the other measurement: $20(1.5-1)=10\ \text{N}$. The line's horizontal intercept gives natural length, not zero length.

Find a hanging equilibrium

• A mass $m$ hangs at rest from one vertical string attached above it. Equilibrium 平衡 requires $T=mg$, so its equilibrium extension is $e=mg/k=mgL/\lambda$.
• A $0.6\ \text{kg}$ mass on $k=30\ \dfrac{\text{N}}{\text{m}}$ has $e=0.2\ \text{m}$ when $g=10\ \dfrac{\text{m}}{\text{s}^2}$. If the mass is moving through that position, acceleration is momentarily zero; speed need not be zero.

Release at natural length is not slow loading

• Attach a mass to the lower end at natural length and release it from rest, with no resistance. After a downward extension $x$, energy gives $mgx=\tfrac12mv^2+\tfrac12kx^2$.
• At maximum stretch $v=0$, so the non-zero root is $x=2mg/k=2e$. Maximum speed occurs earlier at $x=e$, where $mg-kx=0$; $v_{\max}^2=g^2m/k$. Slow loading instead ends at $e$ and removes energy from the system.

A horizontal spring can pass natural length

• A $2\ \text{kg}$ particle on a smooth horizontal line is attached to a spring with $k=50\ \dfrac{\text{N}}{\text{m}}$. Release it from rest at extension $0.4\ \text{m}$. Its initial energy is $\tfrac12k(0.4)^2=4\ \text{J}$.
• At natural length the energy is all kinetic: $\tfrac12mv^2=4$, so $v=2\ \dfrac{\text{m}}{\text{s}}$. The particle continues into compression and first stops at $x=-0.4\ \text{m}$, provided no other physical constraint intervenes.

A horizontal string changes model at natural length

• Replace that spring by an elastic string with the same $k$ and initial extension. It has the same speed at natural length, but beyond that point it has no tension and no elastic energy.
• On the smooth horizontal surface the particle then moves at constant velocity until another constraint acts. Do not predict spring-like compression and reversal. Specify geometry before following a particle past its anchor.

Release a compressed vertical spring

• A spring has lower end fixed, natural length $a$, and $\lambda=10mg$. A mass $m$ is attached to its top and released from rest with compression $a/2$. Let $x$ be signed extension upwards. Relative to natural length, total potential energy is $U=5mgx^2/a+mgx$.
• Initially $x=-a/2$ and $U=3mga/4$. At the next rest point, put $z=x/a$: $5z^2+z=3/4$. The roots are $-1/2$ and $3/10$. The first is the initial state; the maximum extension is $3a/10$.

Maximum speed is not a turning point

• In that spring example, $-kx-mg=0$ at $x=-a/10$. Potential energy there is $-mga/20$, so the maximum kinetic energy is $3mga/4+mga/20=4mga/5$.
• Hence $v_{\max}^2=8ag/5$. At maximum extension $3a/10$, the mass is instantaneously at rest but the acceleration is $-4g$, downwards. A turning point 转折点 has $v=0$; it need not have zero resultant force.

Record both end states before writing energy

• Choose one height reference. Write kinetic energy, gravitational potential energy $mgh$, and every active elastic energy at the initial and final positions. With no resistance their total is conserved; with friction include its negative work.
• Use actual lengths to find each extension independently. A string at or below natural length contributes zero energy, not $\lambda(\ell-L)^2/(2L)$. If a path crosses a slack boundary, use the right force law in each interval.

Free fall can come before stretching

• A mass $m$ is released from rest at its anchor, on a vertical string with natural length $a$ and $\lambda=4mg$. It first falls freely through $a$; the string then begins stretching. At later extension $x$, energy gives $mg(a+x)=\tfrac12mv^2+2mgx^2/a$.
• At the lowest point let $z=x/a$. Then $2z^2-z-1=0$, so $z=1$; the negative root is outside the stretched branch. The total drop is $2a$. Maximum speed is at $x=a/4$, giving $v_{\max}^2=9ag/4$.

Removing a mass changes the equilibrium

• A vertical string has natural length $a$ and $\lambda=4mg$. Masses $m$ and $3m$ first hang at equilibrium, so the extension is $a$. Remove $3m$ without an impulse. The remaining mass starts at rest at length $2a$.
• On its taut branch put $z=x/a$ and take height zero at natural length. Energy becomes $\tfrac12mv^2+mga(2z^2-z)=mga$. To find the first position where $v^2=ag$, solve $2z^2-z=1/2$. The valid root gives $\ell=(5+\sqrt5)a/4$; the other root has negative extension.

A rejected root does not describe the slack flight

• At natural length in that example, $v^2=2ag$ upwards. After that the string is slack; gravity alone acts until the next event. The taut-string quadratic cannot describe this phase.
• After rising a further distance $s$, use $v^2=2ag-2gs$. The later position with $v^2=ag$ is $s=a/2$, so the length is $a/2$. The negative-extension root from the old formula gives a different value because it invents compressive string energy.

Resolve along an inclined plane

• A mass $m$ rests on a smooth plane inclined at $\beta$ to horizontal, attached to an anchor upslope by a string along the plane. Normal reaction is $mg\cos\beta$; downslope weight is $mg\sin\beta$.
• Equilibrium requires $kx=mg\sin\beta$. With $m=1\ \text{kg}$, $g=10\ \dfrac{\text{m}}{\text{s}^2}$, $\sin\beta=3/5$ and $k=20\ \dfrac{\text{N}}{\text{m}}$, $x=0.3\ \text{m}$. Friction, if present, must be included with its actual direction.

Follow a string past its anchor

• Use that smooth plane and particle with natural length $1\ \text{m}$. Release from rest $1.5\ \text{m}$ above the anchor. Initially the string pulls downslope; it then goes slack around the anchor and becomes taut again below it. Maximum speed is below the anchor at extension $0.3\ \text{m}$, where tension balances the $6\ \text{N}$ downslope weight.
• Total travel to that point is $1.5+1+0.3=2.8\ \text{m}$. Work–energy gives $\tfrac12mv^2=\tfrac12k(0.5)^2+mg\sin\beta(2.8)-\tfrac12k(0.3)^2=18.4\ \text{J}$. Thus $v_{\max}^2=36.8\ \dfrac{\text{m}^2}{\text{s}^2}$. Before this point the downslope resultant stays positive.

Friction removes mechanical energy

• On the same incline let $\mu=1/4$, and release the particle from natural length below the anchor. For the first downslope motion, normal reaction is $8\ \text{N}$ and friction is $2\ \text{N}$ upslope. Net driving force before elasticity is $6-2=4\ \text{N}$.
• Work–energy gives $\tfrac12mv^2=4x-10x^2$. The first maximum extension is $0.4\ \text{m}$, and maximum speed occurs at $x=0.2\ \text{m}$ with $v_{\max}^2=0.8\ \dfrac{\text{m}^2}{\text{s}^2}$. Recheck friction's direction on any return journey; this equation describes the first descent.

Two opposing strings need two extensions

• Fix anchors $6\ \text{m}$ apart on a smooth horizontal line. Both strings have natural length $2\ \text{m}$; the left has $\lambda=20\ \text{N}$ and the right $40\ \text{N}$. At coordinate $q$ from the left, both are taut only when $2<q<4$.
• Balance $10(q-2)=20(4-q)$ to find equilibrium $q=10/3\ \text{m}$. Total elastic energy is $U=5(q-2)^2+10(4-q)^2$. A $2\ \text{kg}$ mass released from $q=3\ \text{m}$ has $U=15\ \text{J}$; at equilibrium $U=40/3\ \text{J}$, giving $v_{\max}^2=5/3\ \dfrac{\text{m}^2}{\text{s}^2}$. The other turn is $q=11/3\ \text{m}$, still in the same branch.

Series strings carry a common tension

• For two strings in series 串联 with a light unloaded joining point, tensions are equal and extensions add. Use $x_i=T/k_i$ for each. The equivalent stiffness satisfies $1/k_{\rm eq}=1/k_1+1/k_2$.
• Natural lengths $1,2\ \text{m}$ and moduli $20,30\ \text{N}$ give $k_1=20$, $k_2=15\ \dfrac{\text{N}}{\text{m}}$. Under $6\ \text{N}$ tension their extensions are $0.3,0.4\ \text{m}$. Total extension is $0.7\ \text{m}$; energy is $\tfrac12T(x_1+x_2)=2.1\ \text{J}$.

Parallel springs share an extension

• Two springs in parallel 并联 between the same moving and fixed supports have the same extension if their natural lengths and geometry match. Their forces add: $T=(k_1+k_2)x$, so $k_{\rm eq}=k_1+k_2$.
• For $k_1=20$ and $k_2=30\ \dfrac{\text{N}}{\text{m}}$ at $x=0.2\ \text{m}$, the forces are $4$ and $6\ \text{N}$; total force is $10\ \text{N}$ and total energy $1\ \text{J}$. Equal extensions do not require equal tensions.

A lower string pulls downwards

• Two vertical anchors $A$ above $B$ are $14a$ apart. A particle of mass $\kappa m$ is joined to $A$ by a string of natural length $2a$, and to $B$ by one of natural length $4a$. Both moduli are $4mg$. Let $BP=ya$; then upper and lower extensions are $(12-y)a$ and $(y-4)a$.
• Both are taut for $4<y<12$. Their tensions are $2mg(12-y)$ upwards and $mg(y-4)$ downwards. Equilibrium gives $2(12-y)-(y-4)-\kappa=0$, so $y=(28-\kappa)/3$. Do not add both tensions in the same direction.

Two rest positions can determine the mass

• Release the particle from rest at $y=12$, and suppose its first lower turning point is $y=4$. Take gravitational potential zero at $B$. Total potential energy divided by $mga$ is $V=(12-y)^2+\tfrac12(y-4)^2+\kappa y$.
• The endpoints give $32+12\kappa=64+4\kappa$, hence $\kappa=4$. Between them both extensions are nonnegative. With $\kappa=4$, $V-80=\tfrac32(y-4)(y-12)\leq0$, so the intermediate kinetic energy is feasible.

Use both strings to find maximum speed

• With $\kappa=4$, equilibrium in that example is $y=8$. At release the total energy is $80mga$; at $y=8$ the potential energy is $(16+8+32)mga=56mga$.
• Therefore $\tfrac12(4m)v_{\max}^2=24mga$, giving $v_{\max}^2=12ag$. Use the actual moving mass $4m$ in kinetic energy; the symbol $m$ in the moduli is only the reference mass.

An elastic conical pendulum stretches

• A mass $0.6\ \text{kg}$ rotates steadily with a string of natural length $1.5\ \text{m}$ and modulus $30\ \text{N}$. Let its angle to vertical have $\cos\theta=4/5$, $\sin\theta=3/5$, and take $g=10\ \dfrac{\text{m}}{\text{s}^2}$. Vertical balance gives $T\cos\theta=mg$, hence $T=7.5\ \text{N}$.
• Hooke's law gives $x=TL/\lambda=0.375\ \text{m}$, so actual length $\ell=1.875\ \text{m}$ and orbit radius $r=\ell\sin\theta=1.125\ \text{m}$. The radial equation $T\sin\theta=mr\omega^2$ gives $\omega^2=20/3\ \dfrac{\text{rad}^2}{\text{s}^2}$. Use stretched length in the geometry.

A positive length need not give a possible orbit

• For a non-zero elastic conical orbit, $T=m\ell\omega^2$ and $T=\lambda(\ell-L)/L$. Therefore $\ell=\lambda L/(\lambda-mL\omega^2)$. Its denominator must be positive. Vertical balance also needs $\ell\cos\theta=g/\omega^2$, with $0<\cos\theta<1$.
• Take $L=1\ \text{m}$, $\lambda=20\ \text{N}$, $m=1\ \text{kg}$ and $g=10\ \dfrac{\text{m}}{\text{s}^2}$. If $\omega^2=5$, $\ell=4/3\ \text{m}$ but the required vertical height is $2\ \text{m}$: impossible. If $\omega^2=10$, $\ell=2\ \text{m}$ and $r=\sqrt3\ \text{m}$: possible. At $\omega^2=20$ no finite length solves the equations.

Select a physically valid event

• Mark the initial state, natural-length boundaries, equilibrium positions and candidate turning points. State the active model in each interval: a taut string pulls, a slack string has no force, and a spring may compress.
• Find roots, then check lengths, nonnegative string tensions, nonnegative kinetic energy and the order of events. Maximum speed requires zero tangential resultant within a smooth branch, or comparison with a boundary; instantaneous rest requires $v=0$.

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