Question

36. The displacement of a particle executing S.H.M. is given by x = a sin(π/6t)+bcos(π/6t)
a = 3 cm and b = 4 cm. Find the amplitude and initial phase of S.H.M.​

Answer 1

Answer:

Given,

Position function, X=Asin(πt+ϕ)......(1)

Velocity function, v=dtdX=Aπcos(πt+ϕ)........(2)

Condition, (at t=0, velocity v=πcms−1 andx=1cm )

From equation (1),   1=Asin(π×0+ϕ)......(3)

From equation (2),  π=Aπcos(π×0+ϕ)........(4)

From (3) and (4)

tanϕ=1⇒ϕ=π/4

Put value in equation (3)

1=Asin(π×0+π/4)=Asin(π/4)

A=2cm

Hence, amplitude is 2cm and phase is 4π