A mass oscillates on a spring with a period T and an amplitude 0.48 cm. The mass is at the equilibrium position x = 0 at t = 0, and is moving in the positive direction.
Where is the mass at the time t = 3T/4?
Where is the mass at the time t = 3T/4?
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Given T=T, A=0.48cm
φ = -π/2 since x=0@t=0
ω = 2π/T
The position of SHM is given by:
x(t) = Acos(ωt+φ)
x(3T/4) = 0.48cos((2π/T)(3T/4)+(-π/2))
x = 0.48cos(π) = -0.48
φ = -π/2 since x=0@t=0
ω = 2π/T
The position of SHM is given by:
x(t) = Acos(ωt+φ)
x(3T/4) = 0.48cos((2π/T)(3T/4)+(-π/2))
x = 0.48cos(π) = -0.48