Q.The displacement of an oscillating particle is given by $x=a\sin\omega t+b\cos\omega t$, where a, b and $\omega$ are constants. Prove that the particle performs a linear S.H.M. with amplitude $A=\sqrt{a^2+b^2}$.

[!TLDR] Writing $a=R\cos\phi$, $b=R\sin\phi$ with $R=\sqrt{a^2+b^2}$, $x=a\sin\omega t+b\cos\omega t=R\sin(\omega t+\phi)$ -- a pure S.H.M. …

Given $x=a\sin\omega t+b\cos\omega t$. Introduce two new constants R and $\phi$ defined by $a=R\cos\phi$ and $b=R\sin\phi$ (always possible for any a, b, by choosing $R=\sqrt{a^2+b^2}$ and $\phi=\tan^{-1}(b/a)$). Substituting, $x=(R\cos\phi)\sin\omega t+(R\sin\phi)\cos\omega t=R(\sin\omega t\cos\phi+\cos\omega t\sin\phi)=R\sin(\omega t+\phi)$ using the compound-angle identity $\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi$. This is exactly the standard form of a linear S.H.M., $x=R\sin(\omega t+\phi)$, of amplitude $R=\sqrt{a^2+b^2}$ and initial phase $\phi=\tan^{-1}(b/a)$ -- proving the particle performs linear S.H.M. with amplitude $A=\sqrt{a^2+b^2}$, as …

Introduce $R\cos\phi=a$, $R\sin\phi=b$ and use the compound-angle formula to rewrite the sum as a single sine funct …

Attempting to prove S.H.M. by differentiating twice and checking $\ddot x=-\omega^2x$ without ever identifying the actual amplitu …