Question

1)The displacement of a 0.58 kg mass acted upon by a spring force is given by: x(t)=0.35cos(6.7t) where x is in m and t in s. Show steps

a.Find the velocity of the mass when t = 0.75 s.

b.What is the total mechanical energy when t = 0.75 s?

Answer 1

here displacement , x(t) = 0.35 cos( 6.7t)

by comparing it with general equation which is, x(t) = a cos ( wt)

a( amplitude) = 0.35 ,           w = 6.7 rad/sec

a) V ( velocity) = dx/dt

              = d/dt(0.35 cos ( 6.7t)

= 0.35*( - sin (6.7t) ) * 6.7

at t= 0.75 s

V = - 0.35*6.7*sin( 6.7*0.75)

     = -0.2053 m/s

here -ve sign shows that velocity is towards mean position.

b) total mechanical energy (T.E) = P.E + K.E

P.E = (1/2)*K*x²                                 ( K= spring constant = m*w² )

     = (1/2)*m*w² ( 0.35*cos(6.7 t) )²

    = (1/2)*m*w²*0.35²*cos² ( 6.7t)                               ( here w= 6.7)

K.E = (1/2)*m*V²

= (1/2)*m*( - 0.35*6.7*sin ( 6.7 t) )²

= (1/2)*m*(0.35)²*(6.7)²*sin² (6.7t)

so T.E = P.E+K.E

by putting all the values and rearranging it:

T.E = (1/2)*m*( 6.7)² (0.35)² [ cos² ( 6.7t) + sin² ( 6.7t)]

= (1/2)*0.58*(6.7)²*(0.35)²                                      ( sin²@ + cos²@ = 1)

= 1.594 J