Cyclic frequency of unknown weight

[AdnamaLeigh]

When an unknown weight W was suspended from a spring with an unknown force constant k, it reached its equilibrium position and the spring was stretched 14.2cm because of the weight W. Then the weight W was pulled further down to a position 16cm (1.8cm below its equilibrium position) and released, which caused an oscillation in the spring. Calculate the cyclic frequency of the resulting motion. Answer in Hz.

I set k=F/x:
k=9.8m/.018 = 544.44m

I used T=2π√(m/k) to solve for the period:
T=2π√(m/544.44m) = .269s The mass canceled out

I did 1/T for frequency:
1/.269 = 3.714Hz

The answer is wrong. I think it might have something to do with the 14.2cm information but I don't know how to incorporate it in the problem.

 

[Tide]

You need to use the 14.2 cm to calculate the spring constant.

 

[quicknote]

I would suggest taking a closer look at the given information and equations used to calculate the cyclic frequency. One potential issue could be the use of the incorrect value for the force constant, as it was calculated using the incorrect distance of 14.2cm instead of the actual distance of 1.8cm. Additionally, it may be helpful to consider the fact that the weight was pulled down to a position 1.8cm below its equilibrium position, which could potentially affect the frequency of the resulting motion.

One possible approach to solving this problem could be to first calculate the force constant using the correct distance of 1.8cm, and then using that value to calculate the period and frequency of the oscillation. Another approach could be to consider the change in potential energy of the weight when it is pulled down to the 1.8cm position and released, and how that may affect the resulting motion and frequency.

In general, when faced with a problem like this, it is important to carefully analyze all of the given information and consider any potential factors that may affect the outcome. It may also be helpful to consult with other experts or references in the field to ensure accurate calculations and interpretations.