A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm-1 and oscillates in a damping medium of damping constant 10-2 kg s-1. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to :

Question

A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm-1 and oscillates in a damping medium of damping constant 10-2 kg s-1. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to : (A) 2 s (B) 3.5 s (C) 5 s (D) 7 s

Tardigrade Admin · Accepted Answer

Displacement of this system is given as

< b r / > < b r / > r = A e^{- λ t} 6 s (ω t - δ) < b r / > λ = \frac{R}{2 m} < b r / > R = 1 0^{- 2} k g / s < b r / > λ = \frac{1 0 ^{- 2}}{2 \times 0.1} = 0.05 s^{- 1} < b r / > < b r / >

Energy

= \frac{1}{2} k x^{2} k

is spring constant

< b r / > E = \frac{1}{2} k A^{2} e^{- 2 λ t} cos^{2} (ω t - δ) < b r / >

Suppose

δ = 0

< b r / > < b r / > E_{0} = \frac{K A ^{2}}{2} < b r / > E_{\frac{1}{2}} = \frac{K A ^{2}}{4} = \frac{K A ^{2}}{2} e^{- 2 λ t} cos^{2} (ω t) < b r / > 0.5 = e^{- 0.1 t} cos^{2} (ω t) - < b r / > ω = \frac{m}{k} = \frac{0.1}{640} = 0.0125 < b r / > < b r / >

By putting the value of

ω

in (1) we get,

t ≃ 7

seconds