Question

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 50 cm/s, and if there is no damping, determine the position u of the mass at any time t. (Use g = 9.8 m/s2 for the acceleration due to gravity. Let u(t), measured positive downward, denote the displacement in meters of the mass from its equilibrium position at time t seconds). What does u(t) equal?When does the mass first return to its equilibrium position? (What does t in seconds equal?)

Answer 1

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) = $\dfrac{0.1 \times 9.8}{0.05}$

The equation of the system is expressed as:

$\dfrac{1}{10} u'' + \dfrac{98}{5} u =0$

By estimating the characteristics equation, we have r = ± 14$i$

Thus; the general solution is:

$u(t) = c_1cos \ 14t + c_2 sin 14 \ t$

By applying the initial condition:

u(0) = 0

⇒ 0 = $c_1$

∴

$\dfrac{du}{dt} = ( - c_1 \ sin 14 t )\times 14 +14c_2 \ cos 14 t$

u'(0) = 0.5

0.5 = 14 × c₂

c₂ = 0.5/14

c₂ = 1/28

∴

$\mathbf{u(t) = \dfrac{1}{28} sin 14 t}$

Equating u(t) = 0, we have t = π/14sec as the time when the mass first returns to its equilibrium position.