Answer:
a)0.674 kg b) 2.2 s c) 0.9 m/s²
Explanation:
The amplitude of the ball (xo) = 11.0cm, half way between its equilibrium point its maximum displacement x = 11 cm / 2 = 5.5 cm = 5.5 / 100 in meters = 0.055 meters, speed at this point = 27.2 cm /s = (27.2 / 100) in m/s = 0.272 m/s,
spring constant K = 5.5 N/m
a) The mass of the ball (m) can be calculated using the formula below
v =√ (x²o - x²)K/m
make m subject of the formula
v² = (xo² - x²) K/m
m = K ( xo² - x²) / v²
m = 0.674kg
b) The period of the oscillation can be calculated by the following formula
T = 2π√ (m /K)
substitute the values into the formula
T = 2 × 3.142 × √ (0.674/ 5.5) = 2.2s
c) The maximum acceleration of the ball which occurs at the maximum displacement of the ball can be calculated by the following formula
a = K / m × x ( maximum displacement of the body) = 5.5 / 0.674 × 0.11 = 0.9 m/s²
