Question

The motion of a particle is given by x = A sin ωt + B cos ωt. The motion of the particle is
(a) not simple harmonic
(b) simple harmonic with amplitude A + B
(c) simple harmonic with amplitude (A + B)/2
(d) simple harmonic with amplitude $\sqrt{A^2+B^2}.$

Answer 1

(d) simple harmonic with amplitude $\sqrt{A^2+B^2}$

x = A sin ωt + B cos ωt ...(1)

$$\begin{gathered} \mathrm{Acceleration}, \\ a=\frac{d^{2}x}{d{\mathrm{t}}^2}=\frac{d^2}{d{\mathrm{t}}^2}(A\sin \omega t+B\cos \omega t) \\ =\frac{\mathrm{d}}{\mathrm{dt}}(A\omega \cos \omega t-B\omega \sin \omega t) \\ =-A{\omega }^2\mathrm{sin\omega t}-B{\omega }^2\cos \mathrm{\omega t} \\ =-{\omega }^2(A\sin \omega t\mathit{}+B\cos \omega t) \\ =-{\omega }^{2}x \end{gathered}$$

For a body to undergo simple harmonic motion,
acceleration, a = $-$ kx. ...(2)
Therefore, from the equations (1) and (2), it can be seen that the given body undergoes simple harmonic motion with amplitude, A $=\sqrt{A^2+B^2}$.2