The oscillatory motion of a horizontal spring mass system is described by the fo

Question

The oscillatory motion of a horizontal spring mass system is described by the following equation: x(t) = (0.09m) sin(2 t/(0.92s) + /4). The mass is 1.3kg mass and the spring constant is 36N/m.

(a) Find the maximum speed (in m/s)

(b) Find the position of the mass at t = 0.32s in meters.

(c) Find the speed of the mass at t = 0.32s in m/s

(d) Find the spring potential energy of the system at t = 0.32s in Joules.

(e) Find the mechanical energy of the system relative to the equilibrium position at t = 0.32s in Joules.

Use 10 N/kg for g.

Explanation / Answer

m =1.3 kg , k =36 N/m

x(t) = (0.09m) sin(2 t/(0.92s) + /4)

(a) at t= 0 we get maxium speed

v(t) = (0.09)(2 /(0.92s)

v = 0.09*(2*3.14/0.92)

v =0.61435 m/s

b) at t=0.32

x(t) = (0.09)sin[ (2*3.14*0.32)/(0.92) +/4)

x(t) =0.09 sin[2.9693478]

x(t) = 0.0154 m

c) v(t) =dx/dt

v(t) = (0.09)(2 /(0.92s) cos(2 t/(0.92s) + /4)

v(t) = 0.6053 m/s

c) U =(1/2)kx^2

U = 0.5*36*0.0154*0.0154

U = 0.00427 J

d) K= (1/2)mv^2

K =(0.5*1.3*0.6053*0.6053)

K = 0.2382 J

E = U+K = 0.2382 + 0.00427

E =0.24247 J

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