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.

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.

Damon · Answer

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