Question

A mass of 0.8 kg is fixed at a vertical spring with an unknown spring constant. When spring is released from rest, it extends to 15cm before it stops. As the system oscillates with Simple Harmonic Motion (SHM) the weight is pulled down 20cm further downward and being released.
1. What is the spring constant?
2. What is the total energy of the system?
3. What is the angular frequency of spring?
4. What is the velocity of the system at x=10cm below equilibrium position? ​

Answer 1

Answer:

1) k = 52 N/m

2) E = 1.0 J

3) ω = 8.1 rad/s

4) v = 1.4 m/s

Though asked for a velocity, we can only supply magnitude (speed) because we don't have enough information to determine direction.

If it happens to be the first time it is at y = - 10 cm after release, the velocity is upward.

Explanation:

Assuming the initial setup is after all transients are eliminated.

kx = mg

k = mg/x = 0.8(9.8) / 0.15

k = 52.26666.... ≈ 52 N/m

E = ½kA² = ½(52)(0.20²) = 1.045333... ≈ 1.0 J

ω = √(k/m) = √(52 / 0.8) = 8.0829... ≈ 8.1 rad/s

½mv² = ½kA² - ½kx²

v = √(k(A² - x²)/m) = √(52(0.20² - 0.10²)/0.8) = 1.39999... ≈ 1.4 m/s