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A spring is standing upright on a table with its bottom end fastened to the table. A block is dropped from a height of 3.0 cm above the spring. The block sticks to the top end of the spring and then oscillates with an amplitude of 10 cm. What is the oscillation frequency?

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My first thought was I should be applying energy equations to solve...

[tex] U_{g_{i}} = K_{f} [/tex]

[tex] mgy = 1/2mv^{2} [/tex]

[tex] (9.81m/s^{2})(.03m) = 1/2v^{2} [/tex]

[tex] v_{max} = 0.767 m/s [/tex]

[tex] v_{max} = 2 \pi fA [/tex]

[tex] f = 1.22 Hz [/tex]

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The answer is supposed to be 1.83 Hz..

I'm thinking I'm running into trouble because I made the assumption that the velocity just before the block hits the spring will be the same as the max velocity for the oscillation, but this is the only way I can think to do this question since I don't know the spring constant

Any help would be appreciated

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# Homework Help: Simple Harmonic Motion & a vertical spring

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