P16_021.PDF

(1 / 2 π ) k/m =(1 / 2 π ) g/ . Now the equilibrium point is halfway between the points where the

object is momentarily at rest. One of these points is where the spring is unstretched and the other

is the lowest point, 10cm below. Thus =5 . 0cm=0 . 050m and

f = 1

2 π

9 . 8m / s 2

0 . 050m =2 . 23Hz .

(b) Use conservation of energy. We take the zero of gravitational potential energy to be at the initial

position of the object, where the spring is unstretched. Then both the initial potential and kinetic

energies are zero. We take the y axis to be positive in the downward direction and let y =0 . 080m.

The potential energy when the object is at this point is U = 2 ky 2

−

mgy . The energy equation

becomes 0 = 2 ky 2

−

mgy + 2 mv 2 . We solve for the speed.

v =

2 gy

−

m y 2 = 2 gy

−

y 2

2(9 . 8m / s 2 )(0 . 080m)

−

9 . 8m / s 2

0 . 050m

(0 . 080m) 2 =0 . 56 m / s

ω = k/ ( m + ∆ m ). Thi s should be half the original angular frequency, or 2 k/m .Wesolve

k/ ( m +∆ m )= 2 k/m for m . Square both sides of the equation, then take the reciprocal to

obtain m +∆ m =4 m .Thisgives m =∆ m/ 3 =(300g) / 3 =100g.

(d) The equilibrium position is determined by the balancing of the gravitational and spring forces:

ky =( m +∆ m ) g .Thus y =( m +∆ m ) g/k . We will need to ﬁnd the value of the spring constant

k .Use k = mω 2 = m (2 πf ) 2 .Then

y = ( m +∆ m ) g

m (2 πf ) 2

(0 . 10kg+0 . 30kg) 9 . 8m / s 2

(0 . 10kg)(2 π

2 . 24Hz) 2

=0 . 20m .

This is measured from the initial position.

21. (a) The object oscillates about its equilibrium point, where the downward force of gravity is balanced

by the upward force of the spring. If is the elongation of the spring at equilibrium, then k = mg ,

where k is th e spring con stan t and m is the mass of the object. Thus k/m = g/ and f = ω/ 2 π =