Question

Please solve carefully

6, A damped simple pendulum consists of a bob (m-2.55kg), a length (L = 4m), and a damping force (F- bv). Initially, it oscillates with an amplitude of 16.0 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? (b) How much energy has been "lost" during these four oscillations?

Answer 1

$m\dfrac{d^2 x}{dt^2} +b\dfrac{dx}{dt}+kx = 0$

12

$\gamma = \dfrac{b}{2m}$

$k = \dfrac{mg}{l} \\ \\ T = 2\pi \sqrt{\dfrac{m}{k}} \\ \\ T = 2 \pi \sqrt{\dfrac{l}{g}}$

T = 4.012 sec

after 4 oscillations x(t) = (3/4)A

so damping coefficient $\gamma = \dfrac{b}{2m}$ = 0.0717

so b = 0.366 kg/s

initial energy = $\dfrac{1}{2}k A^2$ = 0.08 Joules

after 4 oscillations energy = $(\dfrac{1}{2}k A^2) e^{-\gamma t}$ = 0.02533 Joules

energy lost = 0.05467 Joules

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