Energy in Simple Harmonic Motion — Question 9

Section 15-3 · Physics pages 2288–2289

Question 9 Equal amounts of extracted energy do not cause equal amplitude drops An otherwise ideal spring oscillator starts with amplitude A0 = .10 m and stiffness k = 100 N/m. A device removes exactly 0.080 J at each successive equilibrium passage while sufficient kinetic energy remains. It supplies no energy between passages. n En 6 Energy after successive equal removals; the last allowed point remains above zero. (a) Find the amplitudes after the first and fifth complete extractions. (b) How many full extractions are possible? State the residual amplitude and explain why extrapolating a linear amplitude decrease is wrong.
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Question 9 - Solution Equal amounts of extracted energy do not cause equal amplitude drops n En 6 (a) Derivation and calculation E0 = 1 2(100)(.10)2 = .500 J, En = .500 −.080n. An = √ 2(.500 −.080n) 100 (En ≥0). A1 = √ .0084 = .09165 m , A5 = √ .0020 = .04472 m . (b) Result and interpretation nmax = ⌊.500 .080 ⌋ = 6 . E6 = .0200 J, A6 = √ .0004 = .0200 m. E7 = −.0600 J < 0. A seventh full removal is impossible; at most the remaining .0200 J can be extracted. Physical check Amplitude decreases as the square root of energy, so equal energy removals do not produce equal amplitude decrements.

Question 9

Equal amounts of extracted energy do not cause equal amplitude drops

An otherwise ideal spring oscillator starts with amplitude A0=.10mA_0=.10\,\mathrm{m} and stiffness k=100N/mk=100\,\mathrm{N/m}. A device removes exactly 0.080J0.080\,\mathrm{J} at each successive equilibrium passage while sufficient kinetic energy remains. It supplies no energy between passages.

Diagram shown in the original SVG worksheet.

  1. Find the amplitudes after the first and fifth complete extractions.

  2. How many full extractions are possible? State the residual amplitude and explain why extrapolating a linear amplitude decrease is wrong.

Question 9 - Solution

Equal amounts of extracted energy do not cause equal amplitude drops

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

E0=12(100)(.10)2=.500J,En=.500−.080n.E_0=\tfrac12(100)(.10)^2=.500\,\mathrm J,\quad E_n=.500-.080n.An=2(.500−.080n)100(En≥0).\boxed{A_n=\sqrt{\frac{2(.500-.080n)}{100}}}\quad(E_n\ge0).A1=.0084=.09165m,A5=.0020=.04472m.A_1=\sqrt{.0084}=\boxed{.09165\,\mathrm m},\quad A_5=\sqrt{.0020}=\boxed{.04472\,\mathrm m}.

(b) Result and interpretation

nmax=⌊.500.080⌋=6.n_{\max}=\left\lfloor\frac{.500}{.080}\right\rfloor=\boxed6.E6=.0200J,A6=.0004=.0200m.E_6=.0200\,\mathrm J,\quad A_6=\sqrt{.0004}=.0200\,\mathrm m.E7=−.0600J<0.E_7=-.0600\,\mathrm J<0.A seventh full removal is impossible; at most the remaining .0200J.0200\,\mathrm J can be extracted.

Physical check

Amplitude decreases as the square root of energy, so equal energy removals do not produce equal amplitude decrements.