Question

Energy (J) 20 PE 15 10 TE 5 0 Tx(cm) 12 16 20 24 28 The energy diagram above describes a mass of 25 grams oscillating in simple harmonic motion on a spring. The spring's equilibrium length is The amplitude of the motion is The spring constant is J/m2 The maximum kinetic energy of the particle is J. It has this kinetic energy as it passes through x = cm.

Answer 1

1. The equilibrium length is X = 20 cm as the potential energy at equilibrium will be zero in a oscillating spring.

2. The system potential energy will be equal to total mechanical energy at the amplitude distance.

Therefore, the amplitude = (26-20) cm = 6cm.

3. Total Energy = 6.5 J.

$\newline TE = \frac{1}{2}KX^{2} \newline \newline 6.5 = \frac{1}{2}K(0.06)^{2} \newline \newline K = 3611.11 J/m^{2}.$

4. The maximum kinetic energy will be at the equilibrium length at X = 20 cm and maximum kinetic energy = 6.5 J which will be same as the total energy.

Thank You !!!