Question

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/mN/m. At t=0 t=0 the block has velocity -4.00 m/sm/s and displacement +0.200 mm.
Find the amplitude.

Answer 1

Answer:

The amplitude of the spring is 32.6 cm.

Explanation:

It is given that,

Mass of the block, m = 2 kg

Force constant of the spring, k = 300 N/m

At t = 0, the velocity of the block, v = -4 m/s

Displacement of the block, x = 0.2 mm = 0.0002 m

We need to find the amplitude of the spring. We know that the velocity in terms of amplitude and the angular velocity is given by :

$v=\omega\sqrt{A^2-x^2}$

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

$\omega=\sqrt{\dfrac{300}{2}}$    

$\omega=12.24\ rad/s$

So, $\dfrac{v^2}{\omega^2}+x^2=A^2$

$\dfrac{(-4)^2}{(12.24)^2}+(0.0002)^2=A^2$            

A = 0.326 m

or

A = 32.6 cm

So, the amplitude of the spring is 32.6 cm. Hence, this is the required solution.