Question

A 1.50-kg object is attached to a spring and placed on
frictionless, horizontal surface. A horizontal force of 28.0 N is
required to hold the object at rest when it is pulled 0.200 m from
its equilibrium position (the origin of the *x* axis). The
object is now released from rest from this stretched position, and
it subsequently undergoes simple harmonic oscillations.

a.)Find the force constant of the spring.

b.)Find the frequency of the oscillations.

c.)Find the maximum speed of the object.

d.)Where does this maximum speed occur?

e.)Find the maximum acceleration of the object.

f.)Where does the maximum acceleration occur?

g.)Find the total energy of the oscillating system.

h.)Find the speed of the object when its position is equal to one-third of the maximum value.

i.)Find the magnitude of the acceleration of the object when its position is equal to one-third of the maximum value.

Answer #1

Force F=28 N

from

a)

F=kx

k= F/x = 28/0.2 = 140 N/m

b)

angular freq w =sqrt(k/m)=sqrt(140/1.50)=9.66rad/s

freq f=w/2pi =9.66/6.28 =1.53 Hz

c)

from consvn of enrgy

0.5kx^2=0.5mv^2

140*0.2^2 =1.50*v^2

speed v =1.93 m/s

d)

maximum speed occurs at mean

e)

maximum accekleration =w^2*A =9.66^2*0.2 =18.66 m/s^2

f)

maximum acceleration occurs at exterme points

g)

total energy =0.5mw^2A^2 =0.5*1.50*9.66^2*0.2^2=2.8 J

h)

speed at x=0.2/3 =0.0667 m

speed v =w*sqrt(A^2-x^2) =1.82 m/s

i)

a =w^2*x =9.66^2*0.0667n =6.22 m/s^2

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