Springs and Elastic Systems flashcards

$F=kx$: force (N) = spring constant k (N/m) times deflection x (m) from natural length. The force points back toward natural length, written $F=-kx$. Valid only in the elastic range.

$U=\frac{1}{2}k{x}^2$ in joules (k in N/m, x in m). Doubling the deflection gives 4 times the energy. Equivalent form: $U=\frac{F^2}{2k}$.

$\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}$, or $k_{eq}=\frac{k_1{k}_2}{k_1+k_2}$ for two. Each carries the full force and stretches add, so the result is softer than the softest spring.

$k_{eq}=k_1+k_2$. The springs share the same deflection and their forces add, so the combination is stiffer than any single spring; n identical springs give nk.

$T=2\pi \sqrt{\frac{m}{k}}$ (s), with $f=1/T$ in hertz. More mass gives a longer period; a stiffer spring a shorter one. Amplitude and gravity do not change T.

The deflection must be in meters: 4 cm = 0.04 m, so $F=500\times 0.04=20$ N. Unconverted centimeters make the answer 100 times too big.

From the natural (unloaded) length, not the hanging equilibrium or installed position. A spring 30 cm relaxed and 34 cm loaded has 4 cm extension, not 34 cm.

Each half has half the coils, so it is twice as stiff (2k). More active coils, thinner wire or a larger coil diameter all make a spring softer.