Question
A mass attached to a spring executes simple harmonic motion in a horizontal plane with an amplitude of 1.07 m. At a point 0.4815 m away from the equilibrium, the mass has speed 3.87 m/s.
What is the period of oscillation of the
mass? Consider equations for x(t) and v(t)
and use sin2 +cos2 = 1 to calculate !.
Answer in units of s.
i'm a bit confused after using the v(t) and x(t) with the cos and sin.
What is the period of oscillation of the
mass? Consider equations for x(t) and v(t)
and use sin2 +cos2 = 1 to calculate !.
Answer in units of s.
i'm a bit confused after using the v(t) and x(t) with the cos and sin.
Answers
x = 1.07 sin (2 pi t/T)
dx/dt = v = (2 pi/T)(1.07) cos (2 pi t/T)
3.87 = 2 pi (1.07/T) cos(2 pi (1.07/T))
also
.4815 = 1.07 sin (2 pi (1.07)/T))
so
.5759 T = cos(mess)
.45 = sin (mess)
square everything
add
solve for T
dx/dt = v = (2 pi/T)(1.07) cos (2 pi t/T)
3.87 = 2 pi (1.07/T) cos(2 pi (1.07/T))
also
.4815 = 1.07 sin (2 pi (1.07)/T))
so
.5759 T = cos(mess)
.45 = sin (mess)
square everything
add
solve for T
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