Question

What is the maximum elastic potential energy of a simple horizontal mass-spring oscillator whose equation of motion is given by x=(0.300m)sin[(3rad/s)t]? The mass on the end of the spring is 0.700kg .

Answer 1

compare the given equation with

x = A*sin(w*t)

we get

A = 0.3 m

w = 3 rad/s

w = sqrt(k/m)

==> k = w^2*m

= 3^2*0.7

= 6.3 N/m

Umax = 0.5*k*A^2

= 0.5*6.3*0.3^2

= 0.2835 J

Answer 2

The expression for the displacement is

      x = A sin?t

Given expression is

       x=(0.300m)sin[(3rad/s)t]

Comparing the above two expressions, we have

       ? = 3 rad/s

Amplitude, A = 0.300 m

The elastic potential energy stored in the spring mass system is given by

Umax= (1/2)m?²A²  

         = 0.5 * 0.7 * 3^2 * 0.3^2 = 0.3 J

Answer 3

The expression for the displacement is

      x = A sin?t

Given expression is

       x=(0.300m)sin[(3rad/s)t]

Comparing the above two expressions, we have

       ? = 3 rad/s

Amplitude, A = 0.300 m

The elastic potential energy stored in the spring mass system is given by

Umax= (1/2)m?²A²  

         = 0.5 * 0.7 * 3^2 * 0.3^2 = 0.3 J