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Maksim231197 [3]
3 years ago
7

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 pulle

d 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
Physics
1 answer:
Harman [31]3 years ago
5 0

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

\omega=\sqrt{\dfrac{k}{m}}

Put the value into the formula

\omega=\sqrt{\dfrac{330}{0.12}}

\omega=5\sqrt{110}\ Hz

We need to calculate the velocity of the block

Using simple harmonic motion

x=A\sin(\omega t+\phi)

x=A\cos\omega t

On differentiating with respect to t

\dfrac{dx}{dt}=-A\omega\sin\omega t

v=-A\omega\sin\omega t

Put the value into the formula

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

v=3.5\ m/s

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

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