Answer:
Step-by-step explanation:
Given that:
mass m = 100 g = 0.1 kg
Length of the spring = 5 cm = 0.05 m
The set in motion from the equilibrium position u(0) = 0
The set in motion from its equilibrium position with a downward velocity u'(0) = 50 cm/s = 0.5 m/s
The spring constant (k) = [tex]\dfrac{0.1 \times 9.8}{0.05}[/tex]
The equation of the system is expressed as:
[tex]\dfrac{1}{10} u'' + \dfrac{98}{5} u =0[/tex]
By estimating the characteristics equation, we have r = ± 14[tex]i[/tex]
Thus; the general solution is:
[tex]u(t) = c_1cos \ 14t + c_2 sin 14 \ t[/tex]
By applying the initial condition:
u(0) = 0
⇒ 0 = [tex]c_1[/tex]
∴
[tex]\dfrac{du}{dt} = ( - c_1 \ sin 14 t )\times 14 +14c_2 \ cos 14 t[/tex]
u'(0) = 0.5
0.5 = 14 × c₂
c₂ = 0.5/14
c₂ = 1/28
∴
[tex]\mathbf{u(t) = \dfrac{1}{28} sin 14 t}[/tex]
Equating u(t) = 0, we have t = π/14sec as the time when the mass first returns to its equilibrium position.
