-->

Mechanical Properties Of Fluids

Question
CBSEENPH11019327

A particle in simple harmonic motion is described by displacement function x(t) = A cos(ωt + ϕ) If the initial position of the particle is 1cm and initial velocity is π cm/s, then what is its amplitude and initial phase? Let angular frequency of particle be π per second.

Solution
Given comma

Displacement space function space of space the space particle space is comma space

space space space space space space space space space space space space space space straight x left parenthesis straight t right parenthesis equals space straight A space cos left parenthesis ωt plus straight ϕ right parenthesis space

straight x left parenthesis 0 right parenthesis space equals space 1 space cm space space space space and space straight v left parenthesis 0 right parenthesis space equals space straight pi space cm divided by sec

Differentiating space straight w. straight r. to space straight t comma space

straight v left parenthesis straight t right parenthesis space equals space minus straight A open parentheses straight omega close parentheses space sin space left parenthesis ωt space plus space straight phi right parenthesis space space space space space space space space... space left parenthesis 1 right parenthesis space

At space straight t equals space 0 comma

straight x left parenthesis 0 right parenthesis space equals space straight A space cos space left parenthesis straight ϕ right parenthesis 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 left parenthesis 2 right parenthesis space

straight v left parenthesis 0 right parenthesis space equals space minus Aω space sin space left parenthesis straight ϕ right parenthesis space space space space space space space space space space space space space space space space space space space space... space left parenthesis 3 right parenthesis thin space

Dividing space left parenthesis 3 right parenthesis space by space left parenthesis 2 right parenthesis comma space

space space space space space fraction numerator straight v left parenthesis 0 right parenthesis over denominator straight x space left parenthesis 0 right parenthesis end fraction space equals space minus straight omega space tan space straight ϕ space

rightwards double arrow space straight pi space equals space minus straight omega space tan space straight ϕ space

rightwards double arrow space space tan space straight ϕ space equals space minus 1 space

straight i. straight e. comma space straight ϕ space equals space fraction numerator 3 straight pi over denominator 4 end fraction space or space fraction numerator 7 straight pi over denominator 4 end fraction

Since comma space at space straight t equals space 0 space

straight x left parenthesis 0 right parenthesis space equals space straight A space cos space open parentheses straight ϕ close parentheses space equals space 1 space

rightwards double arrow space cos space straight ϕ space is space positive. space

Therefore comma space straight ϕ space equals space fraction numerator 7 straight pi over denominator 4 end fraction
therefore space straight A space cos open parentheses space fraction numerator 7 straight pi over denominator 4 end fraction close parentheses space equals space 1 space

rightwards double arrow space space space space space straight A space equals space square root of 2 space cm