When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m
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Answer:
A. 
B. 
C. 
Explanation:
Given:
- spring constant,
- mass attached,
A)
for a spring-mass system the frequency is given as:
B)
frequency is given as:
C)
Time period of a simple harmonic motion is given as:
Explanation:
we are not given the pressure change, check yhe question please
