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A 300 g block connected to a light spring with a force constant of k = 3 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 3 cm from equilibrium and released from rest. Find the period of its motion. (Recall that the period, T, and frequency, f, are inverses of each other.)

Respuesta :

Answer:

Time period, T = 1.98 seconds

Explanation:

It is given that,

Mass of the block, m = 300 g = 0.3 kg

Force constant of the spring, k = 3 N/m

Displacement in the block, x = 3 cm

Let T is the period of the motion of the block. The time period of the block is given by :

[tex]T=2\pi \sqrt{\dfrac{m}{k}}[/tex]

[tex]T=2\pi \sqrt{\dfrac{0.3\ kg}{3\ N/m}}[/tex]    

T = 1.98 seconds

So, the period of the motion of the block is 1.98 seconds. Hence, this is the required solution.

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