A 2kg mass attached to a spring on a horizontal plane with a

A 2kg mass attached to a spring on a horizontal plane with a spring constant of 10 N/m. if the mass is displaced by 3m from its equilibrium position and left to oscillate. Write an equation of motion for the oscillator considering a) no friction between the mass and the surface and b) with friction that is proportional to -2v. c) Find the maximum speed and total energy in both a) and b) cases.

Solution

solution:

1)here without friction equation of motion is given by just inertia and spring force as follows for displacement of x m

mx\'\'+Kx=0

where total energy of system is sumof potential energy in spring and kinetic energy of mass,

total energy=.5*m*x\'^2+.5+K*x^2

for maximum speed

K.E.=P.E.

.5*m*x\'^2=.5*K*x^2

on putting value we get that

x\'^2=.5*10*9/.5*2

x\'=6.7082 m/s

hence total energy by above relation on putting value as

T.E.=P.E.+K.E.=.5*2*6.7082^2+.5*10*9

T.E.=90 N m or j

2)when friction consider thenequation of motion is

mx\'\'+Kx+2x\'=0

here friction force=2*v=2*x\'

where total energy of system is sumof potential energy in spring and kinetic energy of mass and energy loss due to friction,

work done=F*x=2x\'*x

total energy=.5*m*x\'^2+.5+K*x^2+2x\'*x

for maximum speed

K.E.+work done=P.E.

.5*m*x\'^2+2*x\'*x=.5*K*x^2

on putting value we get that

x\'^2+6x\'-45=0

we get that

x\'=4.3484 m/s

hence total energy by above relation on putting value as

T.E.=P.E.+K.E.+work done=.5*2*6.7082^2+.5*10*9+2*4.3484*3

T.E.=90 N m or j

3)hence in both cases total energy of system remain same but just sink of energy transfer changes

.5*m*x^2