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Mechanical Properties Of Fluids

Question
CBSEENPH11019234

A body of mass 0.2kg executes simple harmonic motion of period 4s. The body acquires the velocity 4m/s after 0.6s it passes through its mean position. Find the potential energy of particle at t = 1s.

Solution
The velocity of body at any instant is given by, 
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight v equals Aω space cos space ωt
therefore space space straight A equals fraction numerator straight v over denominator straight omega space cos space ωt end fraction
Time period of oscillation of body = 4s.
Therefore at t = 1s, the body will be at extreme position.
Potential energy stored in the body at t = 1s is,
straight U equals 1 half m omega squared straight A squared equals 1 half m omega squared open parentheses fraction numerator straight v over denominator straight omega space c o s space omega t end fraction close parentheses squared
space space space
space space equals 1 half straight m open parentheses fraction numerator straight v over denominator c o s space omega t end fraction close parentheses squared
We have,
Velocity of the body, v=4 m/s at t=0.6s
Mass of the body, m=0.2kg

a n d space space space space space straight omega equals fraction numerator 2 straight pi over denominator straight T end fraction equals fraction numerator 2 straight pi over denominator 4 end fraction equals straight pi over 2
therefore space space space space space space straight U equals 1 half cross times 0.2 open parentheses fraction numerator 4 over denominator c o s space 0.3 straight pi end fraction close parentheses squared

space space space space space space space space space space space space equals 4.631 straight J