Question

A mass of 0.1kg oscillates in a simple harmonic motion with the amplitude 0.2m and a period of 1.0 Calculate:a. The force constant of the spring b. Maximum potential energy of the massc. The potential energy if the mass is 0.12m from the equilibrium position (π=3.142)​

O a. k = 4π²
O a. k = 25
O a. k = 10
O a. k = 16
O b. 0.0025 J
O b. 0.025 J
O b. 0.00025 J
O b. 0.25 J
O c. 0.012 J
O c. 0.12 J
O c. 0.0012 J
O c. 0.0012 J"

Answer 1

The force constant of the spring is approximately 40 N/m, the maximum potential energy is 0.8 J, and the potential energy at 0.12m from the equilibrium position is 0.288 J.

Step-by-step explanation:

To solve for the force constant of the spring (k), maximum potential energy (Umax), and the potential energy at a distance of 0.12m from the equilibrium position (U), we can use the formulas of Simple Harmonic Motion (SHM). The force constant of the spring (k) can be calculated using the period (T) of the oscillation and the mass (m).

The formula to calculate the force constant is:

k = (4π²m) / T²

Plugging in the values:

k = (4π² × 0.1 kg) / (1.0 s)² = 39.4784 N/m ≈ 40 N/m

The maximum potential energy (Umax) of the mass when it is at the amplitude can be found using the formula:

Umax = ½ kA²

Umax = ½ × 40 N/m × (0.2 m)² = 0.8 J

Finally, the potential energy (U) at 0.12m from the equilibrium can be calculated using the same formula with the new displacement (x):

U = ½ kx²

U = ½ × 40 N/m × (0.12 m)² = 0.288 J