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Question:

A block of mass 0.1kg is connected to an elastic spring of spring constant 640Nm⁻¹ and oscillates in a damping medium of damping constant 10kgs⁻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: 7s, 5s, 3.5s, 2s

7s

2s

3.5s

5s

Solution:

Displacement of this system is given as
r=Ae⁻ᵇt/2m cos(ωt-δ)
λ=R/2m
R=10 kg/s
λ=10/2×0.1=0.05s⁻¹
Energy=1/2kx²
k is spring constant
E=1/2kA²e⁻ᵇt cos²(ωt-δ)
Suppose δ=0
E₀=kA²/2
E/2=kA²/4=kA²/2e⁻ᵇt cos²(ωt)
0.5=e⁻⁰⋅¹t cos²(ωt) ———(1)
ω=√(k/m)=√(640/0.1)=80
By putting the value of ω in (1) we get,
t≈7 seconds