Solve Spring Frequency Homework: .88kg to .4kg

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To solve the problem of finding the new frequency of a spring when the mass is decreased from 0.88 kg to 0.4 kg, the relevant equation is f = (1/2π)√(k/m). The original frequency is 1.8 Hz with a mass of 0.88 kg, leading to the calculation of spring constant k as 112.56 N/m. When the mass is reduced to 0.4 kg, the new frequency can be calculated using the same spring constant. The confusion arises from the different methods used to derive k, but they should yield consistent results. The final frequency for the 0.4 kg mass is approximately 2.7 Hz.
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Homework Statement


A mass of .88 kg when fastened to the lower end of a light-weight spring and set vibrating up and down is found to have a frequency of 1.8 Hz. Calculate the new frequency of vibration when the mass is decreased to .4 kg.

Homework Equations


P = 2 \pi \sqrt{m/k}

The Attempt at a Solution


Using K = (m 4 \pi^2 ) / P^2, my answer for K comes out to 112.56.

But, using \sqrt{k} = (\sqrt{m} 2 \pi )/ P, I get 3.2572. But how can that be? The equations are the same, aren't they? Where am I making the mistake?

Thanks.
 
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2\pi f=\sqrt{\frac{k}{m}}
 
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