Answer:
The spring constant of the spring is 47.62 N/m
Explanation:
Given that,
Mass that is attached with the spring, m = 29 g = 0.029 kg
The spring makes 20 complete vibrations in 3.1 s. We need to find the spring constant of the spring. We know that the number of oscillations per unit time is called frequency of an object. So,

f = 6.45 Hz
The frequency of oscillator is given by :

k is the spring constant


k = 47.62 N/m
So, the spring constant of the spring is 47.62 N/m. Hence, this is the required solution.
Answer:
statement B is true
Explanation:
since same force is applied by the compressed spring on both masses so their rate of change of momenta must be same and since the lighter block has lesser mass so it must have greater velocity to have an equal change in momentum as of heavier mass.
By relation:
, 
comparing momenta of above two equations we get
KElighter (2) = KEheavier (4)
KElighter = 2 KEheavier
The answer is d hope this helps