A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring
and allowed to come to rest. It is then pulled down to a point where the spring is 7.0 cm
longer than its unstretched length, then released. If the ball makes 20 oscillations in 40 s.
(a) Find mass of the ball and its maximum speed?
(b) How much time does it take for the ball to come back to a point 1.0 cm away from its
release point?
(a) To find the mass of the ball and its maximum speed, we can use the equation for the period of oscillation of a mass-spring system:
T = 2π * √(m/k)
Where T is the period, m is the mass, and k is the spring constant.
Given that the ball makes 20 oscillations in 40 seconds, we can calculate the period as:
T = 40 s / 20 = 2 s
Substituting this into the equation, we have:
2 s = 2π * √(m/15 N/m)
Simplifying the equation, we get:
1 = π * √(m/15)
Squaring both sides, we get:
1 = π^2 * m/15
Solving for m, we have:
m = 15 / π^2
m ≈ 1.52 kg
To find the maximum speed, we can use the equation:
v_max = A * 2π / T
Where v_max is the maximum speed and A is the amplitude (the distance the ball is pulled down from the equilibrium position).
Given that the spring is pulled down 7.0 cm longer than its unstretched length, the amplitude is 0.07 m.
Substituting the values into the equation, we have:
v_max = 0.07 m * 2π / 2 s
Simplifying, we get:
v_max = 0.07π m/s
(b) To find the time it takes for the ball to come back to a point 1.0 cm away from its release point, we can use the equation for the time period T:
T = 2π * √(m/k)
Substituting the given values, we have:
T = 2π * √(1.52 kg / 15 N/m)
Simplifying, we get:
T ≈ 2.30 s
Therefore, it takes approximately 2.30 seconds for the ball to come back to a point 1.0 cm away from its release point.