A block of mass 1.6 kg is attached to a spring with a spring constant of 1000 N m⁻¹, as shown in the figure in book 7.14. The spring is compressed through a distance of 2 cm and the block is released from rest. Calculate the velocity of the block as it passes through the equilibrium position, x = 0,  if the surface is frictionless. (Ans: 0.50 m s⁻¹)






Data Given:


Mass of block attached to spring = m = 1.6 kg

Spring constant = k = 1000 N m⁻¹

Amplitude = x₀ = 2 cm = 0.02 m

Displacement at mean position = x = o m


To Find:


The velocity of the mass at mean position = v = ?

Solution:


Since the general formula to find the speed of the body attached to the spring is

v = ധ `\sqrt {x₀^2 - x^2}`

where ധ = `\sqrt frac {k}{m}`

At mean position x = 0

v = `\sqrt frac {k}{m}` x₀

by putting values 

v = `\sqrt frac {1000 N m⁻¹}{1.6 kg}` ｘ 0.02 m

v = `\sqrt {625 s⁻²}` ｘ 0.02 m

v = 25 s⁻¹ｘ 0.02 m

or

v = 0.50 m s⁻¹-------------------------Ans.



************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.