Question

A spring with a spring constant of 1200 N/mhas a 55-gball at its end. The energy of the system is 9.0 J .

A. What is the amplitude A of vibration?

B. What is the maximum speed of the ball?

C. What is the speed when the ball is at a position x=+A/2?

Answer 1

Spring constant = k = 1200 N/m

Mass of the ball = m = 55 g = 0.055 kg

Total energy of the system = E = 9 J

Amplitude of motion = A

The total energy is equal to the maximum potential energy of the spring.

E = kA²/2

9 = (1200)A²/2

A = 0.122 m

Maximum speed of the ball = V_max

The total energy is equal to the maximum kinetic energy of the ball.

E = mV_max²/2

9 = (0.055)V_max²/2

V_max = 18.1 m/s

Displacement of the ball = X = A/2 = 0.122/2 = 0.061 m

Speed of the ball when it is at A/2 = V

The total energy is equal to the sum of the potential energy of the spring and the kinetic energy of the ball.

E = kX²/2 + mV²/2

9 = (1200)(0.061)² + (0.055)V²/2

V = 15.7 m/s

A) Amplitude of vibration = 0.122 m

B) Maximum speed of the ball = 18.1 m/s

C) Speed when the ball is at a position X=+A/2 = 15.7 m/s