Question

A 5.00 kg object on a hor-

izontal frictionless surface is at-

tached to a spring with

k

1000 N/m. The object is dis-

placed from equilibrium 50.0

cm horizontally and given an initial velocity of 10.0 m/s back

toward the equilibrium position. What are (a) the motion’s

frequency, (b) the initial potential energy of the block – spring

system, (c) the initial kinetic energy, and (d) the motion’s am-

plitude?

Answer 1

a) ω = square root(k/m) ; ω = square root (1000/5.00) = 14.14 rad/s

ω=2πf

f=ω/(2π)

f=14.14/6.28

=f = 2.25 Hz

b) potential energy = 0.5kx^2 ; (0.5)(1000)(0.5)² (0.5 being the 50cm it was initially displaced converted to meters) calculating that gives 125J

c) kinetic energy = 0.5mv² ; (0.5)(5.00)(10.0)^2 = 250 J

d) We know that the total energy of the system is given by 0.5kA² and must equal the sum of kinetic and potential energy at any instant. Adding the results we just got for potential and kinetic energy gives us 486J. Let's call this E.

E= 0.5kA² ; square root (2E/k) = A ;

A = square root[(2)(486)/1000] = 0.986 m

I hope help you !!