A box of mass m = 4 kg is suspended vertically from a spring with stiffness k = 64 N/m. Determine the position of the box as a function of time if at t = 0, x = 0 and = 2 m/s.

Respuesta :

Answer:

x=0.5 sin 4 t

Explanation:

Given that:

mass m = 4 kg

Stiffness K =64 N/m

Given spring mass system will be in simple harmonic motion.We know that in simple harmonic motion the natural frequency given as follows

[tex]\omega _n=\sqrt{\dfrac{K}{m}}[/tex]

Now by putting the values

[tex]\omega _n=\sqrt{\dfrac{64}{4}}[/tex]

[tex]\omega _n=4 rad/s[/tex]

The equation of SHM given as

[tex]\ddot{x}+\omega _n^2x=0[/tex]

The solution of above equation will be

[tex]x=Asin\omega _nt[/tex]

x=A sin 4 t

Given at t=0 ,V= 2 m/s

So

V= 4 A cos 4 t

2 = 4 A

A= 0.5

The equation  of motion will be

x=0.5 sin 4 t