Answer:
a) x = 1.5 *10⁻⁴cos(524πt) m
b) v = -1.5 *10⁻⁴(524π)sin(524πt) m/s
a = -1.5 *10⁻⁴(524π)²cos(524πt) m/s²
c) x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm
x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm
Explanation:
x = Acos(ωt)
ω = 2πf = 2π(262) = 524π rad/s
x = 1.5 *10⁻⁴cos(524πt)
v = y' = -Aωsin(ωt)
v = -1.5 *10⁻⁴(524π)sin(524πt)
a = v' = -Aω²cos(ωt)
a = -1.5 *10⁻⁴(524π)²cos(524πt)
not sure about the last part as time is generally not given in mm
I will show at 1 second and at 0.001 s to try to cover bases
x(1) = 1.5 *10⁻⁴cos(524π(1))
x(1) = 1.5 *10⁻⁴cos(524π)
x(1) = 1.5 *10⁻⁴(1)
x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm
x(0.001) = 1.5 *10⁻⁴cos(524π(0.001))
x(0.001) = 1.5 *10⁻⁴cos(0.524π)
x(0.001) = 1.5 *10⁻⁴(-0.0753268)
x(0.001) = -1.129902...*10⁻⁵ m
x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm