+2 votes
in Amplitude by
edited by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
+1 vote
by

To solve this problem, we can use the equations of simple harmonic motion.

  1. Period (T): The period is the time taken for one complete oscillation. It can be calculated using the formula:

T = 2π√(m/k)

Where: m = mass of the object (6.0 kg) k = force constant of the spring (486 Nm^-1)

Plugging in the values, we get:

T = 2π√(6.0/486) ≈ 0.773 seconds

Therefore, the period of oscillation is approximately 0.773 seconds.

  1. Maximum speed (V_max): The maximum speed occurs when the object passes through the equilibrium position. It can be calculated using the formula:

V_max = Aω

Where: A = amplitude (36 cm = 0.36 m) ω = angular frequency, given by ω = √(k/m)

Plugging in the values, we get:

V_max = 0.36 * √(486/6.0) ≈ 3.78 m/s

Therefore, the maximum speed is approximately 3.78 m/s.

  1. Speed at 30 cm displacement (V): To find the speed when the displacement from equilibrium is 30 cm (0.30 m), we can use the equation:

V = ω√(A^2 - x^2)

Where: A = amplitude (0.36 m) x = displacement from equilibrium (0.30 m) ω = angular frequency, given by ω = √(k/m)

Plugging in the values, we get:

V = √(486/6.0) * √(0.36^2 - 0.30^2) ≈ 2.66 m/s

Therefore, the speed when the displacement from equilibrium is 30 cm is approximately 2.66 m/s.

Welcome to Physicsgurus Q&A, where you can ask questions and receive answers from other members of the community.
...