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

TOTAL ENERGY = 0.74 j

Step-by-step explanation:

Given data:

spring constant is 350 N/m

m = 0.24 kg

b = 0.41 kg/s

A = 0.075 M

`\omega = \sqrt{(k)/(m)}`

`= \sqrt{(350)/(0.24)}`

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

`=e^{(-0.41)/(2*0.24) t} cos (\sqrt{(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^{(-0.41)/(2*0.24) 2\pi \sqrt{(0.24)/(350)}} cos (2\pi*0.075)`

y = 0.06517 m

y = 6.517 cm

total energy
`= (1)/(2) k y^2`

`= (1)/(2)* 350* 0.06517^2 = 0.742 J`