Physics · Ch 10 — Oscillations

Vertical oscillations of a spring

10.4.2

Vertical oscillations of a spring

Now hang a massless spring of stiffness constant $k$ from a rigid support (a ceiling), with natural (unloaded) length $L$ . When a mass $m$ is attached to the free end, the spring stretches by an extension $l$ to reach a new equilibrium; at this equilibrium, the upward restoring force of the spring $F_1$ exactly balances the downward weight, $F_1+mg=0$ . Since $F_1=-kl$ (Hooke's law), this gives $mg=kl$ , or $m/k=l/g$ . Now suppose the mass is pulled down a further small distance $y$ and released: the total spring extension is now $(y+l)$ , so the new restoring force is $F_2=-k(y+l)=-ky-kl$ . Applying Newton's second law to the mass, which accelerates as $d^2y/dt^2$ , and using the equilibrium condition $mg=kl$ to cancel the static $kl$ and $mg$ terms, the net equation of motion simplifies to $m\dfrac{d^2y}{dt^2}=-ky$ -- identical in form to the horizontal case, even though gravity is acting throughout. So the time period is again $T=2\pi\sqrt{m/k}$ , exactly matching the horizontal spring's period; gravity shifts the equilibrium point but does not change the period of oscillation about it. Using $m/k=l/g$ , the period can also be written purely in terms …

Figure 10.14Springs

What this figure shows. A short illustrative panel of coil springs of different stiffness, shown side by side to visually introduce the idea that springs differ in how much force is needed to stretch or compress them by a given amount, before the vertical spring-mass system is analysed in detail in the following figure. It sets up the everyday, tangible object -- a coil spring -- that stiffness …

Figure 10.15A massless spring with stiffness constant k

What this figure shows. A spring of natural (unloaded) length L hangs from a fixed ceiling support; when a block of mass m is attached to its lower end the spring stretches by an extra length l, reaching a new equilibrium length L + l. A free-body diagram alongside shows the two forces on the hanging mass at equilibrium -- the upward spring restoring force F1 = -kl and the downward weight Fg = mg -- which exactly balance, and a further panel shows the mass displaced an additional small distance y below this new equilibr …

Misc Example 10.8Spring balance on an unknown planet

Worked out. A spring balance reading 0 to 25 kg has a scale 0.25 m long, taken to planet X where g = 11.5 m/s^2; a mass M is hung on it and oscillates with period 0.50 s, and the task is to find the gravitational force on the body. First the spring's stiffness constant is found from its full-scale reading, k = (25 kg)(11.5 m/s^2)/0.25 m = 1150 N/m, using the fact that the full 25 kg reading corresponds to the full 0.25 m scale extension. Then, since the period of vertical oscillation is T = 2 pi sqrt(M/k), M is found by rearranging to M = k T^2/(4 pi^2) = (1150)(0.50)^2/(4 pi^2) is about 7.3 kg. Finally the gravitational force (weight) on the body on …

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