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A 0.805-kg block oscillates on the end of a spring whose spring constant is k = 40.0 N/m . The mass moves in a fluid which offers a resistive force F=−bv, where b = 0.600 N⋅s/m What is the period of the motion?

Answer :

Explanation:

It is given that,

Mass of the block, m = 0.805 kg

Spring constant of a spring, k = 40 N/m

The mass moves in a fluid which offers a resistive force F = −b v, where b = 0.600 N⋅s/m

The angular frequency of oscillation is given by :

[tex]\omega=\sqrt{\dfrac{k}{m}-(\dfrac{b}{2m})^2}[/tex]

[tex]\omega=\sqrt{\dfrac{40\ N/m}{0.805\ kg}-(\dfrac{0.6\ N.s/m}{2(0.805\ kg)})^2}[/tex]

[tex]\omega=7.03\ rad/s[/tex]

The period of motion is T which is given by :

[tex]T=\dfrac{2\pi}{\omega}[/tex]

[tex]T=\dfrac{2\pi}{7.03}[/tex]

T = 0.89 seconds

So, the period of the motion of the block is 0.89 seconds. Hence, this is the required solution.

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