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A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back toward equilibrium. What is the amplitude A, of the motion? Enter your answer in units of cm but do not include units

Respuesta :

Answer:

7.1 cm.

Explanation:

Given,

mass of the block, m = 5 Kg

Spring constant, k = 2000 N/m

moved position, x = 5 cm

initial speed,v = 1 m/s

Amplitude,A = ?

We know,

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

[tex]\omega = \sqrt{\dfrac{2000}{5}}[/tex]

[tex]\omega = 20\ rad/s[/tex]

Relation between velocity, Amplitude is given by

[tex]v = \omega \sqrt{A^2-x^2}[/tex]

[tex]1 = 20\times \sqrt{A^2-0.05^2}[/tex]

[tex]0.05^2+0.05^2= A^2[/tex]

[tex]A =7.1\ cm[/tex]

Amplitude of the motion is equal to 7.1 cm.

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