The amplitude is 0.383 m and the phase angle is 1.02 rad.
The amplitude is the maximum displacement from the mean position.
Given is the mass m= 2kg and force constant k = 300 N/m, then the angular frequency will be
ω = √(k/m)
Putting the values, we get
ω =√(300/2) = √150 rad/s.
The general form of SHM equation is y = Acos(ωt +Ф)
differentiating w.r.t to time , we get the velocity, v = -Aωsin(ωt +Ф)
At t=0, y₀ = AcosФ and v₀ = -AωsinФ
Given, y₀ =0.2 m and v₀ = -4m/s
then, AcosФ =0.2...............(1)
and
AωsinФ=4 or AsinФ=4/√150...............(2)
Squaring and adding both the equation, we get the Amplitude.
A = 0.383 m
Dividing both the equations, we get the phase angle,
tan Ф = 4/(0.2 x√150) =20/√150
Ф=1.02 rad
Thus, the amplitude is 0.383 m and the phase angle is 0.383m and 1.02 rad.
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