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A 0.545-kg ball is hung vertically from a spring. The spring stretches by 3.56 cm from its natural length when the ball is hanging at equilibrium. A child comes along and pulls the ball down an additional 5cm, then lets go. How long (in seconds) will it take the ball to swing up and down exactly 4 times, making 4 complete oscillations before again hitting its lowest position

Answer :

Answer:

t = 9.52 s

Explanation:

This is an oscillatory motion exercise, in which the angular velocity is

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

Let's use hooke's law to find the spring constant, let's write the equilibrium equation

        F_e - W = 0

        F_e = W

        k x = m g

        k = [tex]\frac{m g}{x}[/tex]

        k = 0.545 9.8 /0.0356

        k = 150 N / m

now the angular velocity is related to the period

          W = 2Ï€ / T

we substitute

          4π² T² = k /m

          T = 4pi² [tex]\sqrt{ \frac{m}{k} }[/tex]

we substitute

           T = 4 pi² [tex]\sqrt{ \frac{0.545}{150} }[/tex]

            T = 2.38 s

therefore for the spring to oscillate 4 complete periods the time is

            t = 4 T

            t = 4 2.38

            t = 9.52 s

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