Question

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

Answer 1

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.