Question

A body is attached to a spring, displaced and then released from rest with an angular frequency of 0.25 rad/s. Determine when the potential energy will be equal to the kinetic energy?

a) T=π/0.25 seconds
b) T=2π/0.25 seconds
c) T=3π/0.25 seconds
d) T=4π/0.25 seconds

Answer 1

The potential energy will be equal to the kinetic energy at a quarter of the period in an oscillating spring system, which can be calculated as T=π/0.25 seconds.

Step-by-step explanation:

In an oscillating spring system, the potential energy (U) and kinetic energy (K) are equal at the points in the motion where the speed of the mass is equal to the average speed over a full cycle. In this system, this occurs at the quarter and three-quarter points of the periodic motion, or at T/4 and 3T/4, where T is the period of the motion. Considering the angular frequency (ω) of 0.25 rad/s in the question, the period (T) is given by T = 2π/ω. To find when the potential energy equals the kinetic energy, we simply divide the period by 4, which results in the answer T=π/0.25 seconds.