Answer to Question #274945 in Physics for jay

Question

A damped harmonic oscillator consists of a block (m = 2.00 kg), aspring (k =

10.0 N/m) and a damping force F = −bv. Initially, it oscillates with an amplitude of

25.0 cm; because of the damping, the amplitude falls to three-fourths of this initial

value at the completion of four oscillations.

a) What is the value of b? (ANS = 0.102 kg/s)

b) How much energy has been lost during these four oscillations?(ANS=0.137 J)

Answer 1

a) $A=A_0e^{-\beta t}\to (3/4)A_0=A_0e^{-\beta 4T}$

$T\approx2\pi\sqrt{m/k}$ So, we get $\ln(4/3)=4\cdot(b/2m)\cdot 2\pi\sqrt{m/k}\to$

$b=\sqrt{mk}\cdot\ln(4/3)/(4\pi)=\sqrt{2\cdot10}\cdot\ln(4/3)/(4\cdot3.14)=0.102\ (kg/s)$

b) Initial energy $E_0=kx_0^2/2=10\cdot0.25^2/2=0.3125\ (J)$

Final energy $E=k(3x_0/4)^2/2=10\cdot(3\cdot0.25/4)^2/2=0.1758\ (J)$

$\Delta E=0.1758-0.3125=-0.137\ (J)$