A spring, whose spring constant is 80.0 N m⁻¹ vertically supports a mass of 1.0 kg is at rest position. Find the distance by which the mass must be pulled down, so that on being released, it may pass the mean position with a velocity of one metre per second. (x₀ = 0.11 m)



Given:

Spring constant = k = 80 N m⁻¹

Mass attached to spring = m = 1 kg

The velocity of the mass at mean position = v = 1 m s⁻¹


To Find:


Amplitude = x₀ = ?

Solution:


Since we know 

v = ധ x₀

where ധ = `\sqrt frac {k}{m}`

so 

v = `\sqrt frac {k}{m}` x₀

or

x₀ = `\sqrt frac {m}{k}` v

by putting values 

x₀ = `\sqrt frac {1 kg}{80 N m⁻¹}` ｘ 1 m s⁻¹ 

or

x₀ = 0.112 m -------------------------Ans.



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