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ch4aika [34]
3 years ago
7

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.
Physics
1 answer:
tia_tia [17]3 years ago
3 0

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

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