Answer to Question #100460 in Physics for Sarada mandal

Question

A particle is subjected simultaneously to two simple harmonic
motions in the same direction each of same frequency but of
different amplitudes and a phase difference Find the
amplitude of the resultant displacement and its phase relation to
one of the components.

Answer 1

Let

$$x_1=A_1 \cos(\omega t), \quad x_2=A_2 \cos(\omega t+\phi)$$

Hence

$$x=x_1+x_2=A_1 \cos(\omega t)+A_2 \cos(\omega t+\phi)$$

$$=A_1 \cos(\omega t)+A_2 \cos(\omega t)\cos(\phi)-A_2 \sin(\omega t)\sin(\phi)$$

$$= \cos(\omega t)(A_1+A_2\cos(\phi))-\sin(\omega t)(A_2 \sin(\phi))$$

$$=A\cos(\omega t+\varphi)$$

where

$$A\cos(\varphi)=A_1+A_2\cos(\phi),\; A\sin(\varphi)=A_2\sin(\phi)$$

Hence

$$A=\sqrt{(A_1+A_2\cos\phi)^2+(A_2\sin \phi )^2}$$

$$\tan\varphi=\frac{A_2\sin\phi}{A_1+A_2\cos\phi}$$