Задача 31234 Решить все ...
Условие
Решение
F(t)=m*a(t)
a(t)=v`(t)=(s`(t))`
s`(t)=(5π/3)*cos((π/3)t+(π/6))
a(t)=(5π^2/9)*(-sin((π/3)t+(π/6)));
a(t)=(-5π^2/9)sin((π/3)t+(π/6))
F(0)=m*(5π^2/9)*(-sin((π/3)*0+(π/6)))= -5π^2m/18
4.
1)12sin2t+5cos2t=13*((12/13)*sin2t+(5/13)cos2t)=
=13*(sin θ sin2t+cos θ cos2t)=13*cos(2t-θ )
sin θ =12/13; cos θ =5/13
2)8*sin(5x+(π/6))-15cos(5x+(π/6))=17*((8/17)*sin(5x+(π/6))-(15/17)*cos(5x+(π/6)))=17*(sin θ sin(5x+(π/6))-cosθ*cos(5x+(π/6)))
= - 17*cos(5x+(π/6)+θ)
sin θ =8/17; cos θ =15/17
5.
1)
y=y_(1)+y_(2)=3sin(t/2)+5sin(t/2)=8sin(t/2)=8*cos((π/2)-(t/2))=
=8*cos((t/2)-(π/2))
A=8 - амплитуда
θ= -π/2 - начальная фаза
2)
y=y_(1)+y_(2)=2*sin2t+2sin(2t+(π/3))=2*(sin2t+sin(2t+(π/3))=
=2*2sin((2t+2t+(π/3))/2)*cos((2t-2t-(π/3))/2)=
=4sin(2t+(π/6))*cos(-π/6)=
=2sqrt(3)*sin((2t+(π/6))=
=2sqrt(3)cos((π/2)-(2t+(π/6)))=
=2sqrt(3)cos((π/2)-2t-(π/6))=
=2sqrt(3)cos((2π/6)-2t)=
=2sqrt(3)cos(2t - (π/3))
A=2sqrt(3) - амплитуда
θ= -π/3 - начальная фаза
3)
y=y_(1)+y_(2)=2sqrt(5)sin5t+sqrt(2)cos5t=
=sqrt(22)*((2sqrt(5)/sqrt(22))*sin5t+(sqrt(2)/sqrt(22)*cos5t)=
=sqrt(22)*(sin θ*sin5t+cosθ*cos5t)=
=sqrt(22)*cos(5t-θ)
A=sqrt(22) - амплитуда
θ= - начальная фаза
sin θ =2sqrt(5/22); cos θ =sqrt(2/22)
Применяли метод введения вспомогательного угла
(см. приложения)
и формулы тригонометрии
cos( α + β )=
cos ( α - β )=
формулы приведения
sin γ =cos((π/2)- γ )