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3.35 Solution of simple harmonic motion

Let

f⁢(x)=A⁢cos⁡β⁢x+Bβ⁢sin⁡β⁢xf(x)=A\cos\beta x+{{B}\over{\beta}}\sin\beta x

then we have

f′⁢(x)=-β⁢A⁢sin⁡β⁢x+B⁢cos⁡β⁢xf^{\prime}(x)=-\beta A\sin\beta x+B\cos\beta x
f′′⁢(x)=-β2⁢A⁢cos⁡β⁢x-B⁢β⁢sin⁡β⁢xf^{\prime\prime}(x)=-\beta^{2}A\cos\beta x-{{B}{\beta}}\sin\beta x

so f⁢(0)=Af(0)=A, f′⁢(0)=Bf^{\prime}(0)=B and

f′′⁢(x)=-β2⁢f⁢(x)=-km⁢f⁢(x).f^{\prime\prime}(x)=-\beta^{2}f(x)=-{{k}\over{m}}f(x).

Also, ff is periodic since

f⁢(x+2⁢π/β)=A⁢cos⁡β⁢(x+2⁢π/β)+Bβ⁢sin⁡β⁢(x+2⁢π/β)f(x+2\pi/\beta)=A\cos\beta(x+2\pi/\beta)+{{B}\over{\beta}}\sin\beta(x+2\pi/\beta)
=Acosβx+Bβsinβx=f(x).=A\cos\beta x+{{B}\over{\beta}}\sin\beta x=f(x).