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## Homework Statement

Prove that the maximum speed (Vmax) of a mass on a spring is given by 2(pi)(f)(A)

## Homework Equations

E(total) = kA^2

Vmax = sqrt[2E(total) / m]

(Because E(potential) = 0 when V is at max, so E(total) = mv^2 / 2 + 0)

f = sqrt(k/m) / 2(pi)

k = [(f)(2pi)]^2(m)

## The Attempt at a Solution

Vmax = sqrt[2(kA^2) / m]

Vmax = sqrt [2[(f)(2pi)]^2 (m) A^2)] / m]

Square the [(f)(2pi)]^2, factor in m, factor in 2A^2 and cancel out all m's then sqrt I get:

2Af2(pi)

Did I make a mistake somewhere? Or did I go about this all wrong? :/