A brief note on Force Law, Springs and Energy in SHM . Read it before a test. Try each question at the end before you open its answer.
The force law Simple harmonic motion needs a restoring force in proportion to the displacement: F = − k x F = -kx F = − k x . Angular frequency: ω = k / m \omega = \sqrt{k/m} ω = k / m . Period: T = 2 π m / k T = 2\pi\sqrt{m/k} T = 2 π m / k , and f = 1 / T f = 1/T f = 1/ T . The period does not depend on the amplitude, or on g g g . Springs Hanging spring: at rest k d = m g kd = mg k d = m g , and T = 2 π d / g T = 2\pi\sqrt{d/g} T = 2 π d / g . Side by side, in parallel: k = k 1 + k 2 k = k_1 + k_2 k = k 1 + k 2 . End to end, in series: 1 k = 1 k 1 + 1 k 2 \dfrac{1}{k} = \dfrac{1}{k_1} + \dfrac{1}{k_2} k 1 = k 1 1 + k 2 1 . A block between two springs and two walls: parallel. A piece of a spring is stiffer: k × k \times k × length stays the same. Two bodies on one spring: use μ = m 1 m 2 m 1 + m 2 \mu = \dfrac{m_1m_2}{m_1 + m_2} μ = m 1 + m 2 m 1 m 2 . Energy Potential energy: U = 1 2 k x 2 U = \tfrac12 kx^2 U = 2 1 k x 2 . Total energy: E = 1 2 k A 2 E = \tfrac12 kA^2 E = 2 1 k A 2 . Speed: v = ω A 2 − x 2 v = \omega\sqrt{A^2 - x^2} v = ω A 2 − x 2 , largest v m a x = ω A v_{max} = \omega A v ma x = ω A at the middle. K = U K = U K = U at x = A / 2 x = A/\sqrt2 x = A / 2 . Each energy repeats every T / 2 T/2 T /2 and averages to E / 2 E/2 E /2 . A worked example A block of 0.30 kg 0.30\ \text{kg} 0.30 kg on a spring of 120 N/m 120\ \text{N/m} 120 N/m swings with an amplitude of 4.0 cm 4.0\ \text{cm} 4.0 cm .
Angular frequency. ω = 120 N/m 0.30 kg = 20 rad/s \omega = \sqrt{\dfrac{120\ \text{N/m}}{0.30\ \text{kg}}} = 20\ \text{rad/s} ω = 0.30 kg 120 N/m = 20 rad/s . Period. T = 2 π 20 rad/s = 0.314 s T = \dfrac{2\pi}{20\ \text{rad/s}} = 0.314\ \text{s} T = 20 rad/s 2 π = 0.314 s . Energy. E = 1 2 × 120 N/m × ( 0.040 m ) 2 = 0.096 J E = \tfrac12 \times 120\ \text{N/m} \times (0.040\ \text{m})^2 = 0.096\ \text{J} E = 2 1 × 120 N/m × ( 0.040 m ) 2 = 0.096 J . Largest speed. v m a x = 20 rad/s × 0.040 m = 0.80 m/s v_{max} = 20\ \text{rad/s} \times 0.040\ \text{m} = 0.80\ \text{m/s} v ma x = 20 rad/s × 0.040 m = 0.80 m/s . Practice with new numbers
A program to try The program works through the example above.
Run it. Compare each line with the worked example.
1. Change m = 0.30 to m = 1.20. Which numbers change, and which stays the same?
Mistakes to avoid Writing T = 2 π k / m T = 2\pi\sqrt{k/m} T = 2 π k / m . The mass goes on top. Adding spring constants in series. In series, add the inverses. Forgetting to square the amplitude in 1 2 k A 2 \tfrac12 kA^2 2 1 k A 2 . Putting equal energies at A / 2 A/2 A /2 . They are equal at A / 2 A/\sqrt2 A / 2 . Check yourself A block on a spring has a period of 0.5 s 0.5\ \text{s} 0.5 s . The spring is swapped for one four times as stiff. What is the new period? Stuck? Show a hint Show the answerTwo springs of 30 N/m 30\ \text{N/m} 30 N/m and 60 N/m 60\ \text{N/m} 60 N/m are joined end to end. What is the spring constant of the pair? Stuck? Show a hint Show the answerThe amplitude of an oscillator is halved. What happens to its largest speed and to its energy? Stuck? Show a hint Show the answerWhy does a hanging spring have the same period as the same spring on a smooth floor? Stuck? Show a hint Show the answer