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A 0.12 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 330 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t = 0.7 s is closest to: A) -4.9 m/s B) 4.9 m/s C) 3.5 m/s D) -3.5 m/s E) zero

Respuesta :

Answer:

The velocity of the block is 3.5 m/s.

(C) is correct option.

Explanation:

Given that,

Mass of block = 0.12 kg

Force constant = 330 N/m

Time = 0.7 s

We need to calculate the angular frequency

Using formula of angular frequency

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

Put the value into the formula

[tex]\omega=\sqrt{\dfrac{330}{0.12}}[/tex]

[tex]\omega=5\sqrt{110}\ Hz[/tex]

We need to calculate the velocity of the block

Using simple harmonic motion

[tex]x=A\sin(\omega t+\phi)[/tex]

[tex]x=A\cos\omega t[/tex]

On differentiating with respect to t

[tex]\dfrac{dx}{dt}=-A\omega\sin\omega t[/tex]

[tex]v=-A\omega\sin\omega t[/tex]

Put the value into the formula

[tex]v=-0.080\times5\sqrt{110}\sin(5\sqrt{110}\times0.7)[/tex]

[tex]v=3.5\ m/s[/tex]

Hence, The velocity of the block is 3.5 m/s.

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