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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.

User Undone
by
8.3k points

1 Answer

2 votes

Answer:

The amplitude of the spring is 32.6 cm.

Step-by-step 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√(A^2-x^2)


\omega=\sqrt{(k)/(m)}


\omega=\sqrt{(300)/(2)}


\omega=12.24\ rad/s

So,
(v^2)/(\omega^2)+x^2=A^2


((-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.

User Chris Brandt
by
7.9k points
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