Question

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

Answer 1

7.1 cm.

Step-by-step 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,

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

`\omega = \sqrt{(2000)/(5)}`

`\omega = 20\ rad/s`

Relation between velocity, Amplitude is given by

`v = \omega √(A^2-x^2)`

`1 = 20* √(A^2-0.05^2)`

`0.05^2+0.05^2= A^2`

`A =7.1\ cm`

Amplitude of the motion is equal to 7.1 cm.