CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A 2g block attached to an ideal spring with a spring constant of 80N/m  oscillates on a horizontal frictionless surface.When the spring is 4.0 cm shorter than its equilibrium length,the speed of 17^1/2m/s.The greatest speed of the block is 

1)4m/s

2)6m/s

3)5m/s

4)9m/s

5)3m/s


Solution


This is SHM.

k = 80 N/m            m = 0.002kg
ω = angular frequency of SHM of spring mass system
=> ω²  =  k/m = 80/0.002         => ω = 200 rad/sec²

Let the equation of SHM be:    x = x₀  Sin (200 t)
                                    and    v = x₀ * 200  Cos( 200 t)

Given  that  displacement  0.04 meters = x₀ Sin(200 t)
and  speed:                        √17 m/s = x₀ 200 * Cos (200t)

Hence,   0.04² + (√17/200)² = x₀²
                 x₀ = 0.0449m or 4.499cm

Thus the maximum speed of the block = x₀ * ω 
      =0.0449×200
      = 8.98 m/sec
      =9m/s

 

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image