Question

An 8.00 lb weight stretches a certain spring 0.500 ft. With this weight attached, the spring is pulled 3.00 inches longer than its equilibrium length and released. Find the equation of the resulting motion, assuming no damping.

Answer 1

Answer:

x(t) = 0.75 Cos8t

Explanation:

we know

F = m g

k x = m g

k = $\frac{m g}{x}$

k = $\frac{8 \times 32}{0.5}$

k = 512 lb/s²

now,

$m\frac{\mathrm{d}^2x }{\mathrm{d} t^2}+kx=0$

m = 8 lb    k = 512 lb/s²

$8\frac{\mathrm{d}^2x }{\mathrm{d} t^2}+512x=0$

x(t) = C₁Cos8t + C₂Sin8t

at t = 0  x = 0.5 ft + 3 inch = 0.75 ft

C₁ = 0.75 ft

x'(t) = -0.75×8Sin8t + 8C₂Cos8t

at t=0      x'(0) = 0

 8C₂ = 0

hence

x(t) = 0.75 Cos8t