A mass at the end of spring describes S.H.M with T = 0.40 s. Find out 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 = a = ?


Solution:


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

a = - x ധ²

but ധ = 2πT   ⇒   ധ² = 4π2T2 So,

a = - x 4π2T2

by putting values 

a = - 0.04 m x 4(3.1416)2(0.40s)2

a = - 1.5790.16  m s⁻²

a = - 9.869  m s⁻² -------------Ans.



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