A mass at the end of spring describes S.H.M with T = 0.40 s. Find out `\vec a` when the displacement is 0.04 m. (a = -9.86 m s⁻² approx) 




Given:

Time period = T  = 0.40 s

Displacement = x = 0.04 m

To Find:


Acceleration = `\vec a` = ?


Solution:


The relations of linear acceleration 'a' and angular acceleration 'ധ' is

`\vec a` = - x ധ²

but ധ = `\frac {2π}{T}`   ⇒   ധ² = `\frac {4π^2}{T^2}` So,

`\vec a` = - x `\frac {4π^2}{T^2}`

by putting values 

`\vec a` = - 0.04 m ｘ `\frac {4(3.1416)^2}{(0.40 s)^2}`

`\vec a` = - `\frac {1.579}{0.16}`  m s⁻²

`\vec a` = - 9.869  m s⁻² -------------Ans.



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