Equation of Simple Harmonic Motion of Spring Derivation

Hooke's Law states that:

F=kxF+kx=0Fm+kxm=0a+kxm=0d2x(t)dt2+kx(t)m=0\begin{aligned} F&=-kx\\ F+kx&=0\\ \frac{F}{m}+\frac{kx}{m}&=0\\ a+\frac{kx}{m}&=0\\ \frac{d^2x(t)}{dt^2}+\frac{kx(t)}{m}&=0\\ \end{aligned}

Assume x(t)eλtx(t)\propto e^{\lambda t} solves the differential equation:

d2eλtdt2+keλtm=0λ2eλt+keλtm=0eλt(λ2+km)=0λ2+km=0λ2=kmλ=±ikm\begin{aligned} \frac{d^2e^{\lambda t}}{dt^2}+\frac{ke^{\lambda t}}{m}&=0\\ \lambda^2e^{\lambda t}+\frac{ke^{\lambda t}}{m}&=0\\ e^{\lambda t}\left(\lambda^2+\frac{k}{m}\right)&=0\\ \lambda^2+\frac{k}{m}&=0\\ \lambda^2&=-\frac{k}{m}\\ \lambda&=\pm i\sqrt{\frac{k}{m}} \end{aligned}

{x1(t)=c1eikmtx2(t)=c2eikmt\begin{cases} x_1(t)=c_1e^{i\sqrt{\frac{k}{m}} t}\\ x_2(t)=c_2e^{-i\sqrt{\frac{k}{m}} t} \end{cases}

x(t)=x1(t)+x2(t)=c1eikmt+c2eikmt=c1[cos(kmt)+isin(kmt)]+c2[cos(kmt)isin(kmt)]=c1cos(kmt)+c1isin(kmt)+c2cos(kmt)c2isin(kmt)=(c1+c2)cos(kmt)+(c1c2)isin(kmt)=C1cos(kmt)+C2sin(kmt)=C1+C2cos[kmttan1(C2C1)]\begin{aligned} x(t)&=x_1(t)+x_2(t)\\ &=c_1e^{i\sqrt{\frac{k}{m}} t}+c_2e^{-i\sqrt{\frac{k}{m}} t}\\ &=c_1\left[\cos\left(\sqrt{\frac{k}{m}}t\right)+i\sin\left(\sqrt{\frac{k}{m}}t\right)\right]+c_2\left[\cos\left(\sqrt{\frac{k}{m}}t\right)-i\sin\left(\sqrt{\frac{k}{m}}t\right)\right]\\ &=c_1\cos\left(\sqrt{\frac{k}{m}}t\right)+c_1i\sin\left(\sqrt{\frac{k}{m}}t\right)+c_2\cos\left(\sqrt{\frac{k}{m}}t\right)-c_2i\sin\left(\sqrt{\frac{k}{m}}t\right)\\ &=(c_1+c_2)\cos\left(\sqrt{\frac{k}{m}}t\right)+(c_1-c_2)i\sin\left(\sqrt{\frac{k}{m}}t\right)\\ &=C_1\cos\left(\sqrt{\frac{k}{m}}t\right)+C_2\sin\left(\sqrt{\frac{k}{m}}t\right)\\ &=\sqrt{C_1+C_2}\cos\left[\sqrt{\frac{k}{m}}t-\tan^{-1}\left(\frac{C_2}{C_1}\right)\right]\\ \end{aligned}

Let A=C1+C2A=\sqrt{C_1+C_2}, ω2=km\omega^2=\frac{k}{m} and tan(φ)=C2C1\tan(\varphi)=\frac{C_2}{C_1}:

x(t)=Acos(ωtφ)\boxed{x(t)=A\cos\left(\omega t-\varphi\right)}

Proof by differentiation:

x(t)=Acos(ωtφ)dx(t)dt=Aωsin(ωtφ)d2x(t)dt2=Aω2cos(ωtφ)d2x(t)dt2=ω2[Acos(ωtφ)]d2x(t)dt2=kmx(t)d2x(t)dt2+kx(t)m=0\begin{aligned} x(t)&=A\cos\left(\omega t-\varphi\right)\\ \frac{dx(t)}{dt}&=-A\omega\sin\left(\omega t-\varphi\right)\\ \frac{d^2x(t)}{dt^2}&=-A\omega^2\cos\left(\omega t-\varphi\right)\\ \frac{d^2x(t)}{dt^2}&=-\omega^2[A\cos\left(\omega t-\varphi\right)]\\ \frac{d^2x(t)}{dt^2}&=-\frac{k}{m}x(t)\\ \frac{d^2x(t)}{dt^2}+\frac{kx(t)}{m}&=0\quad\blacksquare \end{aligned}

Note by Gandoff Tan
9 months ago

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Nice, all though this has been done so many times that it's a matter of copy-paste from physics forums. However, it's pretty impressive considering you did it without prior knowledge from scratch

Krishna Karthik - 5 months, 1 week ago

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