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 08m) sin(2pi t/(0 92s) + pi/4). The mass is 1.3kg mass and the song constant is 65N/m Find the maximum speed fin m/s) Find the position of the mass at t = 0 37s in meters Find the speed of the mass at t = 0 37s in m/s Find the spring potential energy of the system at t = 0 37s in Joules Find the mechanical energy of the system relative to the equilibrium position at t = 0 37s in Joules Use 10 N/kg for g

Explanation / Answer

              = 0.546 m/s

b) Just use the x equation and plug in t = 0.37
x(0.37) = 0.08 m * sin(2*pi*0.37/0.92+pi/4)

            = 0.0046199 m

c) plug in t = 0.37 into the v equation
v(0) = (2*pi/0.92s)*(0.08m)*cos(2*3.14*0.37/0.92 + pi/4)

      = 0.5450 m/s

d) Using x(0.37), SPE = 1/2*k*x^2

SPE= 0.5*65*(0.0046199)2

       = 5.65*10-4 J

                               = 0.5*1.3*0.54502 +5.65*10-4 J

                                  = 0.193066+5.65*10-4

                                              = 0.19363 J

This should also be equal to 1/2*m*v_max^2     

= 0.19377 J

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