Oscillations of 1.5 kg Block on Spring: Frequency and Stretch

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A 1.5 kg block attached to a spring with a force constant of 300 N/m oscillates with a frequency of 2.3 Hz after being pulled down 2.0 cm and released. The maximum stretch of the spring during oscillations is 6.9 cm, while the minimum stretch is 2.9 cm. The equilibrium position is determined by the weight of the block, and the oscillation occurs around this point. The calculations involve using the spring constant and understanding the relationship between force, displacement, and equilibrium. The discussion clarifies how to approach the problem of finding the maximum and minimum stretch of the spring.
Dooga Blackrazor
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1. A block of mass 1.5 kg is attached to the end of a vertical spring of force constant k=300 N/m. After the block comes to rest, it is pulled down a distance of 2.0 cm and released.
(a) What is the frequency of the resulting oscillations?
(b) What are the maximum and minimum amounts of stretch of the spring during the oscillations of the block?

(a) 2.3 Hz
(b) 2.9 cm and 6.9 cm


I used the formula T = 2(pie)(squaroot of m/k) with f = 1 /T and found the first answer; however, I'm now sure how to calculate the answers for the second part of the question. F = k*x (spring constant times displacement, but I'm not sure how that factors into the two answers.

Thanks in advance.
 
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Dooga Blackrazor said:
I'm now sure how to calculate the answers for the second part of the question. F = k*x (spring constant times displacement, but I'm not sure how that factors into the two answers.
They are measuring the stretch of the spring from its unstretched position (before the mass is attached). First find out where the equilibrium position is when the mass is attached. Then realize that the mass will oscillate about that point with the given amplitude.
 
Thanks - I've got it now.
 
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