A mass at the end of spring describes S.H.M with T = 0.40 s. Find out 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:
Solution:
The relations of linear acceleration 'a' and angular acceleration 'ധ' is
= - x ധ²
but ധ = ⇒ ധ² = So,
= - x
by putting values
= - 0.04 m x
= - m s⁻²
= - 9.869 m s⁻² -------------Ans.
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© 2022-Onwards by Academic Skills and Knowledge (ASK)
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