Question

in damped oscillation mass is 1kg and spring constant = 100N/m^2 , damping coefficient = 0.5 kg s^-1 . if the mass is displaced by 10 cm from mean position than what will be the value of its mechanical energy after 4 sec ?
a} 0.67J. b} 0.067J. c} 6.7 J. d} 0.5J​

Answer 1

Answer: (b). The mechanical energy is 0.067 J.

Explanation:

Given that,

Mass m = 1 kg

Spring constant $k = 100 N/m^{2}$

Damping coefficient $b = 0.5 kg/s$

Distance $x = 10 cm$

Time $t = 4 s$

We know,

The energy for damped oscillation is

$E = \dfrac{1}{2} kx^{2}e^{-\dfrac{bt}{m}}$

$E = \dfrac{1}{2}\times 100\ n/m^{2}\times0.01 m^{2}\times e^{-\dfrac{0.5\times 4}{1}}$

$E = \dfrac{e^{-2}}{2}$

$E = 0.067 J$

Hence, the mechanical energy is 0.067 J.