Question

A 3.00 kg object rests upon a frictionless, horizontal floor. The object is attached to a horizontal spring (of force constant k = 485 N/m) whose other end is anchored to a nearby wall. The object is pulled until it lies a distance xi = 4.85 cm from its equilibrium position (x = 0). The object is then released and undergoes simple harmonic motion.

Answer 1

Time period =0.49

Angular frequency =12.71

Step-by-step explanation:

As this block is released from equilibrium position to a distance of 4.85 cm it's amplitude is 4.85 cm.

For a body which undergoes simple harmonic motion it's angular frequency =
`\sqrt{(k)/(m) }` =
`\sqrt{(480)/(3) }`=12.71

Time period of this block =
`(2\pi )/(angular frequency)=(2\pi )/(12.71)` =0.49 seconds