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

Answer:

Time period =0.49

Angular frequency =12.71

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 =$\sqrt{\frac{k}{m} }$ =$\sqrt{\frac{480}{3} }$=12.71

Time period of this block =$\frac{2\pi }{angular frequency}=\frac{2\pi }{12.71}$ =0.49 seconds