Question

A damped harmonic oscillator consists of a mass on a spring, with a damping force proportional to the speed of the block. If the spring constant is 350 N/m, the mass of the block is 240 g, the damping constant is 0.41 kg/s, and the block is displaced 7.5 cm from its equilibrium position and then released, what is its kinetic energy after one cycle?A. 7500 JB.0.85 JC.0.98 JD.0.74 JE.1.2 s

Answer 1

Answer:

TOTAL ENERGY = 0.74 j

Explanation:

Given data:

spring constant is 350 N/m

m = 0.24 kg

b = 0.41 kg/s

A = 0.075 M

$\omega = \sqrt{\frac{k}{m}}$

             $= \sqrt{\frac{350}{0.24}}$

$y = e^{\frac{-b}{2m} t} A cos(\omega t)$

  $=e^{\frac{-0.41}{2*0.24} t} cos (\sqrt{\frac{350}{0.24}} t) *0.075$

after one full cycle, mass will be at extreme point, hence K,E = 0

But total energy remain same

y after one full cycle is\

$y =e^{\frac{-0.41}{2*0.24} 2\pi \sqrt{\frac{0.24}{350}}} cos (2\pi*0.075)$

y = 0.06517 m

y = 6.517 cm

total energy $= \frac{1}{2} k y^2$

$= \frac{1}{2}* 350* 0.06517^2 = 0.742 J$