"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its speed when it passes the equilibrium position."

Respuesta :

Answer:

1.3m/s

Explanation:

Data given,

Mass,m=1.0kg,

Amplitude,A=0.10m,

Frequency,f=2.0Hz.

From the equation of a simple harmonic motion, the displacement of the object at a given time is define as

[tex]x=Acos\alpha \\[/tex]

we can express the velocity by the derivative of the displacement,

Hence

[tex]V=-Awsin\alpha \\[/tex]

at equilibrium, the velocity becomes

[tex]V=wA\\w=2\pi f[/tex]

Hence if we substitute values we arrive at

[tex]V=2\pi fA\\V=2\pi *2*0.1\\V=1.3m/s[/tex]

Answer:

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Explanation:

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