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Question
Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by x = `5sin((pit)/3)`m.
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Solution 1
Given:
x = `5sin((pit)/3)`m
t = 1s
∴ ν = `(dx)/(dt) = d/dt(5sin((pit)/3)) = 5cos((pit)/3) xx pi/3`
Put t = 1s ...(Given)
∴ ν = `5cos(pi/3) xx pi/3` = 2.6179 m/s
Solution 2
t = 1s, x = 5 sin60
A = 5, x = `2.5sqrt3` ...........(given)
v = `wsqrt(A^2 - x^2)` .........(Formula)
v = `pi/3sqrt(25 - 18.74)`
v = 1.04 × 2.5
v = 2.61 m/s
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