Answer:
Explanation:
Value of spring constant K = weight / extension
=[tex]\frac{32\times32}{2}[/tex]
=512 ft s⁻²
Frequency of oscillation of spring mass system
= [tex]\sqrt{\frac{k}{m} }[/tex]
Putting the values we get
frequency
=[tex][tex]\sqrt{\frac{512}{32} }[/tex][/tex]
= 4 .
Time period = 1 / frequency = .25 s
No of vibrations in 4π second
= [tex]\frac{4\pi }{.25}[/tex]
= 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
