Ametal ball attached to a spring moves in simple harmonic motion. the amplitude of the ball's motion is 11.0 cm, and the spring constant is 5.50 n/m. when the ball is halfway between its equilibrium position and its maximum displacement from equilibrium, its speed is 27.2 cm/s.(a) what is the mass of the ball (in ) what is the period of oscillation (in ) what is the maximum acceleration of the ball? (enter the magnitude in m/s2.)

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

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