Question

Obtain the equation of motion for a 1.25-g mass, at the end of a perfectly elastic spring which, when stretched 3.75 cm from equilibrium and then released from rest, undergoes simple harmonic motion with a period of 0.016667 s. B.) Find the (i) spring constant (ii) maximum velocity and (iii) total energy of the mass. C.) For the above spring find the positions for which the potential energy is one-third the kinetic energy.

Answer 1

Considering the spring mass system ,

Here the only conservative force is acting on the system(Spring force is conservative in nature), therefore the total mechanical energy will be constant .

Let the spring constant be k .

initially the spring is stretched 3.75 cm from the equilibrium and the mechanical energy is conserved for the system.

Let = angular frequency =

period of oscillation is given by

  

(as time period is given 0.016667 s)

sec^-1

177.64 N/m

ii ).

Here the mechanical is conserved for the system i.e K + U = constant thus the kinetic energy will be maximum when the potential is minimum which is zero .This situation will arrive when the object will be at equilibrium position .

initially ,

kinetic energy = 0 J (as the initial velocity is 0 m/s)

potential energy by the spring is given by

initially x = 3.75 cm = 0.0375 m

initial potential energy = J

= 0.125 J

Thus the mechanical energy = 0.125 j

At equilibrium , mechanical energy = 0.125 J

kinetic energy + 0 = 0.125 j

   = 0.125 J

   = 0.125

   m/s = 14.14 m/s

Thus is the velocity of the mass at equilibrium position .

iii ) .

Total energy = 0.125 J

C ) .

Let the kinetic energy be y J therefore the potential energy = y/3 J.

`Mechanical energy = = 0.125 J

= = 0.125 J

  

Potential energy = = 0.03125 J

and potential energy is given by .Therefore ,

  

  

   m

At a distance of 0.018 m from the equilibrium position the mass will have potential energy one-third of kinetic energy .