Exercises — Spring-mass system — horizontal, vertical

Every symbol below was built in the parent note. Quick meanings:

• $m$ = mass (kg), $k$ = spring stiffness / "stiffness" in newtons per metre (N/m).
• $x$ = displacement from equilibrium (the rest point where net force is zero), in metres.
• $\omega =\sqrt{k/m}$ = angular frequency (rad/s) — how many radians of the cosine cycle pass per second.
• $T=2\pi \sqrt{m/k}$ = period (s) = time for one full back-and-forth.
• $f=1/T$ = frequency (Hz) = oscillations per second (Hz).
• $A$ = amplitude (m) = the farthest the mass strays from equilibrium.
• $v_{max}=A\omega$ = top speed, reached at the centre.
• $y_0=mg/k$ = static stretch of a hanging spring; $g=9.8{\text{m/s}}^2$ throughout.

WHAT we do: plug straight into the definitions — this is pure recognition. WHY: these three quantities are just three faces of the same number. $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2.0}}=\sqrt{25}=5.0\text{rad/s}$ $T=\frac{2\pi }{\omega }=\frac{2\pi }{5.0}=1.257\text{s}$ $f=\frac{1}{T}=\frac{1}{1.257}=0.796\text{Hz}$ Sanity check: $2\pi \sqrt{m/k}=2\pi \sqrt{2/50}=2\pi (0.2)=1.257$ s. ✓

WHAT we do: convert to metres, then use the static-stretch shortcut. WHY this tool: we are told neither $m$ nor $k$ — but the parent note showed $T=2\pi \sqrt{y_0/g}$, which needs only the stretch, because $k$ and $m$ hide together inside $y_0=mg/k$. $y_0=2.5\text{cm}=0.025\text{m}$ $T=2\pi \sqrt{\frac{y_0}{g}}=2\pi \sqrt{\frac{0.025}{9.8}}=2\pi \sqrt{2.551\times 10^{-3}}=2\pi (0.05051)=0.317\text{s}$

WHAT: use $\omega =5.0$ rad/s from before. (a) Max speed — occurs at the centre where all energy is kinetic (Energy in SHM): $v_{max}=A\omega =0.15\times 5.0=0.75\text{m/s}$ (b) Max acceleration — occurs at the turning points where displacement is largest. From $\ddot{x}=-{\omega }^{2}x$, the size of acceleration peaks at $x=\pm A$: $a_{max}={\omega }^{2}A=(5.0{)}^2\times 0.15=25\times 0.15=3.75{\text{m/s}}^2$ (c) Total energy — all of it is spring PE at full stretch: $E=\frac{1}{2}k{A}^2=\frac{1}{2}(50)(0.15{)}^2=\frac{1}{2}(50)(0.0225)=0.5625\text{J}$ Cross-check: $\frac{1}{2}m{v}_{max}^2=\frac{1}{2}(2.0)(0.75{)}^2=0.5625$ J. ✓ Two roads, one energy.

WHAT: period first, then invert $T=2\pi \sqrt{m/k}$ to solve for $k$. WHY invert: we know $T$ and $m$, we want $k$; algebra rearranges the same formula. $T=\frac{9.0}{12}=0.75\text{s}$ Square both sides of $T=2\pi \sqrt{m/k}$: $T^2=4{\pi }^2\frac{m}{k}\Rightarrow k=\frac{4{\pi }^{2}m}{T^2}=\frac{4{\pi }^2(0.30)}{(0.75{)}^2}=\frac{11.84}{0.5625}=21.1\text{N/m}$

WHAT (a): at rest the spring force balances gravity, so $k{y}_0=mg$. $k=\frac{mg}{y_0}=\frac{1.2\times 9.8}{0.08}=\frac{11.76}{0.08}=147\text{N/m}$ (b) Period — vertical, but the derivation cancels gravity, so use the horizontal formula: $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{1.2}{147}}=2\pi \sqrt{8.163\times 10^{-3}}=2\pi (0.09035)=0.568\text{s}$ (c) Max speed with $\omega =\sqrt{k/m}=\sqrt{147/1.2}=\sqrt{122.5}=11.07$ rad/s: $v_{max}=A\omega =0.05\times 11.07=0.553\text{m/s}$ (d) Total (net) force — measured from equilibrium, the net restoring force is $F_{net}=-kx$. It is largest in size at the turning points $x=\pm A$, and it points back toward equilibrium, so it takes opposite signs at the two ends. Taking down as positive: $F_{net}=-kx\Rightarrow F_{net}\text{ranges from }-kA\text{ to }+kA$ $F_{net,\text{at bottom }(x=+A)}=-kA=-147\times 0.05=-7.35\text{N (points up, toward equilibrium)}$ $F_{net,\text{at top }(x=-A)}=+kA=+147\times 0.05=+7.35\text{N (points down, toward equilibrium)}$ So the net force ranges from $-7.35$ N to $+7.35$ N (magnitude $7.35$ N at each turning point, zero at equilibrium). If instead you want the spring force alone (which fights gravity), the extension ranges from $y_0-A$ to $y_0+A$: $F_{spring,min}=k(y_0-A)=147(0.03)=4.41\text{N (at the top)}$ $F_{spring,max}=k(y_0+A)=147(0.13)=19.11\text{N (at the bottom)}$

WHAT: same spring means same $k$, so $T=2\pi \sqrt{m/k}\propto \sqrt{m}$ — the constant $2\pi /\sqrt{k}$ cancels in a ratio. WHY a ratio: we never need $k$; dividing the two periods deletes everything except the masses. $\frac{T_2}{T_1}=\frac{2\pi \sqrt{m_2/k}}{2\pi \sqrt{m_1/k}}=\sqrt{\frac{m_2}{m_1}}=\sqrt{\frac{0.90}{0.40}}=\sqrt{2.25}=1.5=\frac{3}{2}$ Then: $T_2=1.5\times T_1=1.5\times 0.50=0.75\text{s}$

WHAT/WHY: the block loses contact the instant the surface can no longer pull it down — i.e. the surface's own downward acceleration at the top equals gravity. Contact is lost when the peak downward acceleration reaches $g$. The largest acceleration in SHM is $a_{max}={\omega }^{2}A$ and it points toward equilibrium; at the top it points down. Setting it equal to $g$: ${\omega }^{2}A=g\Rightarrow \omega =\sqrt{\frac{g}{A}}=\sqrt{\frac{9.8}{0.04}}=\sqrt{245}=15.65\text{rad/s}$ $f=\frac{\omega }{2\pi }=\frac{15.65}{2\pi }=2.49\text{Hz}$ Beautiful point: the mass $M$ cancels — it never appears. Any block leaves at the same frequency, because both the required and available forces scale with $M$.

WHAT: the total energy equals its full-stretch value $\frac{1}{2}k{A}^2$. Solve for $v$. $\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2$ Multiply by $2/m$ and use ${\omega }^2=k/m$: $v^2=\frac{k}{m}(A^2-x^2)={\omega }^2(A^2-x^2)$ $\boxed{v(x)=\pm \omega \sqrt{A^2-x^2}}$ Check the ends: at $x=0$, $v=\omega \sqrt{A^2}=A\omega =v_{max}$ ✓ (fastest at centre). At $x=\pm A$, $v=\omega \sqrt{A^2-A^2}=0$ ✓ (momentarily still at the turning points). The $\pm$ reflects that the mass passes each interior point once going each way.

Which formula for a stretch-only vertical problem? ::: $T=2\pi \sqrt{y_0/g}$ Where is speed maximum, where is acceleration maximum? ::: Speed max at centre ($x=0$); acceleration max at ends ($x=\pm A$). At what $x$ is KE = PE? ::: $x=\pm A/\sqrt{2}$ Parallel vs series effective stiffness? ::: Parallel $k_1+k_2$ (stiffer); series $1/k_s=1/k_1+1/k_2$ (softer). How do you fix the phase constant sign? ::: Use the sign of $v(0)$; position alone is ambiguous. Speed as function of position? ::: $v=\omega \sqrt{A^2-x^2}$