A 0.63-kg metal sphere oscillates at the end of a vertical spring. As the spring

Question

A 0.63-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to 0.21 m (relative to its unstrained length), the speed of the sphere decreases from 5.8 to 4.6 m/s. What is the spring constant of the spring?

Explanation / Answer

The velocity/position relation in SHM is v² = (k/m)*(A² - x²) (see below) 5.7² = (k/m)*(A² - 0.16²) 4.0² = (k/m)*(A² - 0.25²) find A² from the first A² - 0.16² = 5.7² *(m/k) A² = 5.7² *(m/k) + 0.16² put it in the second 4.0² = (k/m)*([5.7² *(m/k) + 0.16²] - 0.25²) 16.0 = 32.49 + 0.0256*(k/m) - 0.0625*(k/m) -16.49 = -0.0369*(k/m) k/m = 446.8 k = 416 N/m --------------------------------------… NOTE: Derivation of velocity/distance relation m*d²x/dt² = -k*x ----> d²x/dt² = dv/dt m*dv/dt = -k*x ----> dv/dt = (dv/dx)*(dx/dt) = v*dv/dt v*dv = -(k/m)*x*dx integrate this to get v²/2 = -(k/m)*x²/2 + C v = 0 at x = max amplitude A 0 = -(k/m)*A²/2 + C so C = (k/m)*A²/2 v²/2 = -(k/m)*x²/2 + (k/m)*A²/2 v² = (k/m)*(A² - x²)

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