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What's the velocity in this spring problem?

A 0.50-kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.15m. What is the velocity velocity when it passes the equilibrium point?

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Now, I know how to get the answer. From the question, I can just get the distance from the diameter, and the time to cover that distance from the frequency. With that method, I get my book's answer, which is 2.8m/s.

The question is, how come I can't just find the spring constant from the formula

f=(1/2pi)(k/m)^0.5 k is the spring constant, pi is 3.14, m is mass, f is frequency

...then plug in the constant into the potential energy equation, put amplitude in as the distance, set potential energy equal to kinetic energy, and find velocity? How come it's a different answer?Let:
U be the potential energy at displacement x,
K be the kinetic energy at displacement x,
w be angular frequency,
m be the mass.

U = mw^2 x^2 / 2
K = mw^2(a^2 - x^2) / 2

The sum U + K is constant.
At the extremity of the motion, x = a, K = 0, and U = mw^2 a^2 / 2.

The equilibrium point is at the centre of oscillation, where x = 0.
At that point the K = mw^2 a^2 / 2, and U = 0.

You appear to have U and K confused.