A) a 32-pound weight stretches a spring 2 feet. determine the amplitude and period of motion if the weight is released 1 foot above the equilibrium position with an initial upward velocity of 2 ft/sec. (b) how many complete vibrations will the weight have completed at the end of 4π seconds?

Explanation:

Value of spring constant K = weight / extension

=

=512 ft s⁻²

Frequency of oscillation of spring mass system

=

Putting the values we get

frequency

=" alt="\sqrt{\frac{512}{32} }" />" />

= 4  .

Time period = 1 / frequency = .25 s

No of vibrations in 4π second

=

= 50 approx

When the weight is released 1 foot above equilibrium

PE stored = 1/2 K x² = .5 x 512 x 1 = 256 J

KE = 1/2 mv² = .5 x 32 x 4 = 64 J

Total energy =  320 J

Let x be the amplitude upto which spring stretches due to this energy

1/2 k x² = 320 ,

1/2 x 512  x² = 320

x = 1.12 ft approx

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//