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A 0.50-kg box is attached to an ideal spring of force constant (spring constant) 20 N/m on a horizontal, frictionless floor. The box oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. (a) What is the amplitude of vibration

Respuesta :

Answer:

[tex]A = 0.24 m[/tex]

Explanation:

As we know that for spring block system the angular frequency is given as

[tex]\omega = \sqrt{\frac{k}{m}}[/tex]

now we know that

[tex]k = 20 N/m[/tex]

also we know that

[tex]m = 0.50 kg[/tex]

so we have

[tex]\omega = \sqrt{\frac{20}{0.5}}[/tex]

[tex]\omega = 6.32 rad/s[/tex]

now we know that speed of an SHM at its equilibrium position is given as

[tex]v = A\omega[/tex]

now plug in all values in it

[tex]1.5 = A(6.32)[/tex]

[tex]A = 0.24 m[/tex]

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