A 0.52-kg metal sphere oscillates at the end of a vertical spring. As the spring

Question

A 0.52-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.13 to 0.21 m (relative to its unstrained length), the speed of the sphere decreases from 5.2 to 4.3 m/s. What is the spring constant of the spring?

Explanation / Answer

mass m = 0.52 kg speed at x= 0.13 m is v = 5.2 m / s w *sqrt [ A^ 2- x^ 2] = 5.2 ----( 1) speed at x '= 0.21 m is v = 4.3 m / s w *sqrt [ A^ 2- x ' ^ 2] = 4.3 ----( 2) eq( 1) / eq( 2) ==> [A^ 2-x^ 2] / [A^ 2-x'^ 2] = ( 5.2/4.3)^ 2 = 1.4624 A^ 2-0.13^ 2= 1.4624 [ A^ 2- 0.21^ 2] A^ 2- 0.0169 = 1.4624A^ 2- 0.0533] from this A ^ 2 = 0.0787 A = 0.28 m plug this in eq( 1) we get angular frequency w value we know spring constant of the spring k = mw^ 2 plug the values we get answer

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