Composition of two S.H.M.s having same period and along the same path

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   i. Consider a particle simultaneously subjected to two S.H.M.s having the same period and along same path (let it be along the x-axis), but of different amplitudes and initial phases. The resultant displacement at any instant is equal to the vector sum of its displacements due to both the S.H.M.s at that instant. ii. Let the two linear S.H.M's be given by equations, $x_1=A_1\sin(\omega t+\phi_1)\qquad....(1)$ $x_2=A_2\sin(\omega t+\phi_2)\qquad....(2)$ where $A_1$, $A_2$ are amplitudes; $\phi_1$, $\phi_2$ are initial phase angles and $x_1$, $x_2$ are the displacement of two S.H.M's in time 't'. ω is same for both S.H.M's. iii. The resultant displacement of the two S.H.M's is given by, $x=x_1+x_2\qquad....(3)$ iv. Using equations (1) and (2), equation (3) can be written as, $x=A_1\sin(\omega t+\phi_1)+A_2\sin(\omega t+\phi_2)$ $=A_1[\sin\omega t\cos\phi_1+\cos\omega t\sin\phi_1]+A_2[\sin\omega t\cos\phi_2+\cos\omega t\sin\phi_2]$ $=A_1\sin\omega t\cos\phi_1+A_1\cos\omega t\sin\phi_1+A_2\sin\omega t\cos\phi_2+A_2\cos\omega t\sin\phi_2$ $=[A_1\sin\omega t\cos\phi_1+A_2\sin\omega t\cos\phi_2]+[A_1\cos\omega t\sin\phi_1+A_2\cos\omega t\sin\phi_2]$ $\therefore x=\sin\omega t[A_1\cos\phi_1+A_2\cos\phi_2]+\cos\omega t[A_1\sin\phi_1+A_2\sin\phi_2]\qquad....(4)$ v. As $A_1$, $A_2$, $\phi_1$ and $\phi_2$ are constants, we can combine them in terms of another convenient constants R and δ as $A_1\cos\phi_1+A_2\cos\phi_2=R\cos\delta\qquad....(5)$ and $A_1\sin\phi_1+A_2\sin\phi_2=R\sin\delta\qquad....(6)$ vi. Using equations (5) and (6), equation (4) can be written as, $x=\sin\omega t.\,R\cos\delta+\cos\omega t.\,R\sin\delta=R[\sin\omega t\cos\delta+\cos\omega t\sin\delta]$ $\therefore x=R\sin(\omega t+\delta)\qquad....(7)$ Equation (7) is the equation of an S.H.M. of the same angular frequency (hence, the same period) but of amplitude R and initial phase δ. It shows that the combination (superposition) of two linear S.H.M.s of the same period and occurring along the same path is also an S.H.M. Resultant Amplitude: Resultant amplitude is, $R=\sqrt{(R\sin\delta)^2+(R\cos\delta)^2}$ Squaring and adding equations (5) and (6) we get, $(A_1\cos\phi_1+A_2\cos\phi_2)^2+(A_1\sin\phi_1+A_2\sin\phi_2)^2=R^2\cos^2\delta+R^2\sin^2\delta$ $\therefore A_1^2\cos^2\phi_1+A_2^2\cos^2\phi_2+2A_1A_2\cos\phi_1\cos\phi_2+A_1^2\sin^2\phi_1+A_2^2\sin^2\phi_2+2A_1A_2\sin\phi_1\sin\phi_2=R^2(\cos^2\delta+\sin^2\delta)$ $\therefore A_1^2(\cos^2\phi_1+\sin^2\phi_1)+A_2^2(\cos^2\phi_2+\sin^2\phi_2)+2A_1A_2(\cos\phi_1\cos\phi_2+\sin\phi_1\sin\phi_2)=R^2$ $\therefore A_1^2+A_2^2+2A_1A_2\cos(\phi_1-\phi_2)=R^2$ $\therefore R=\pm\sqrt{A_1^2+A_2^2+2A_1A_2\cos(\phi_1-\phi_2)}\qquad....(8)$ Equation (8) represents resultant amplitude of two S.H.M's. Initial Phase: Dividing equation (6) by (5), we get, $\dfrac{A_1\sin\phi_1+A_2\sin\phi_2}{A_1\cos\phi_1+A_2\cos\phi_2}=\dfrac{R\sin\delta}{R\cos\delta}=\tan\delta$ $\therefore \delta=\tan^{-1}\left[\dfrac{A_1\sin\phi_1+A_2\sin\phi_2}{A_1\cos\phi_1+A_2\cos\phi_2}\right]\qquad....(9)$ Equation (9) represents resultant or initial phase of two S.H.M's.

2. 2.↑ back to the question

   [From 5.10, Q1 (except Initial Phase)]

   i. Consider a particle simultaneously subjected to two S.H.M.s having the same period and along same path (let it be along the x-axis), but of different amplitudes and initial phases. The resultant displacement at any instant is equal to the vector sum of its displacements due to both the S.H.M.s at that instant. ii. Let the two linear S.H.M's be given by equations, $x_1=A_1\sin(\omega t+\phi_1)\qquad....(1)$ $x_2=A_2\sin(\omega t+\phi_2)\qquad....(2)$ where $A_1$, $A_2$ are amplitudes; $\phi_1$, $\phi_2$ are initial phase angles and $x_1$, $x_2$ are the displacement of two S.H.M's in time 't'. ω is same for both S.H.M's. iii. The resultant displacement of the two S.H.M's is given by, $x=x_1+x_2\qquad....(3)$ iv. Using equations (1) and (2), equation (3) can be written as, $x=A_1\sin(\omega t+\phi_1)+A_2\sin(\omega t+\phi_2)$ $=A_1[\sin\omega t\cos\phi_1+\cos\omega t\sin\phi_1]+A_2[\sin\omega t\cos\phi_2+\cos\omega t\sin\phi_2]$ $=A_1\sin\omega t\cos\phi_1+A_1\cos\omega t\sin\phi_1+A_2\sin\omega t\cos\phi_2+A_2\cos\omega t\sin\phi_2$ $=[A_1\sin\omega t\cos\phi_1+A_2\sin\omega t\cos\phi_2]+[A_1\cos\omega t\sin\phi_1+A_2\cos\omega t\sin\phi_2]$ $\therefore x=\sin\omega t[A_1\cos\phi_1+A_2\cos\phi_2]+\cos\omega t[A_1\sin\phi_1+A_2\sin\phi_2]\qquad....(4)$ v. As $A_1$, $A_2$, $\phi_1$ and $\phi_2$ are constants, we can combine them in terms of another convenient constants R and δ as $A_1\cos\phi_1+A_2\cos\phi_2=R\cos\delta\qquad....(5)$ and $A_1\sin\phi_1+A_2\sin\phi_2=R\sin\delta\qquad....(6)$ vi. Using equations (5) and (6), equation (4) can be written as, $x=\sin\omega t.\,R\cos\delta+\cos\omega t.\,R\sin\delta=R[\sin\omega t\cos\delta+\cos\omega t\sin\delta]$ $\therefore x=R\sin(\omega t+\delta)\qquad....(7)$ Equation (7) is the equation of an S.H.M. of the same angular frequency (hence, the same period) but of amplitude R and initial phase δ. It shows that the combination (superposition) of two linear S.H.M.s of the same period and occurring along the same path is also an S.H.M. Resultant Amplitude: Resultant amplitude is, $R=\sqrt{(R\sin\delta)^2+(R\cos\delta)^2}$ Squaring and adding equations (5) and (6) we get, $(A_1\cos\phi_1+A_2\cos\phi_2)^2+(A_1\sin\phi_1+A_2\sin\phi_2)^2=R^2\cos^2\delta+R^2\sin^2\delta$ $\therefore A_1^2\cos^2\phi_1+A_2^2\cos^2\phi_2+2A_1A_2\cos\phi_1\cos\phi_2+A_1^2\sin^2\phi_1+A_2^2\sin^2\phi_2+2A_1A_2\sin\phi_1\sin\phi_2=R^2(\cos^2\delta+\sin^2\delta)$ $\therefore A_1^2(\cos^2\phi_1+\sin^2\phi_1)+A_2^2(\cos^2\phi_2+\sin^2\phi_2)+2A_1A_2(\cos\phi_1\cos\phi_2+\sin\phi_1\sin\phi_2)=R^2$ $\therefore A_1^2+A_2^2+2A_1A_2\cos(\phi_1-\phi_2)=R^2$ $\therefore R=\pm\sqrt{A_1^2+A_2^2+2A_1A_2\cos(\phi_1-\phi_2)}\qquad....(8)$ Equation (8) represents resultant amplitude of two S.H.M's. Initial Phase: Dividing equation (6) by (5), we get, $\dfrac{A_1\sin\phi_1+A_2\sin\phi_2}{A_1\cos\phi_1+A_2\cos\phi_2}=\dfrac{R\sin\delta}{R\cos\delta}=\tan\delta$ $\therefore \delta=\tan^{-1}\left[\dfrac{A_1\sin\phi_1+A_2\sin\phi_2}{A_1\cos\phi_1+A_2\cos\phi_2}\right]\qquad....(9)$ Equation (9) represents resultant or initial phase of two S.H.M's. vii. a. When phase difference is zero radians: Two S.H.M.s are in phase, i.e., $\phi_1-\phi_2=0$ $\therefore \cos(\phi_1-\phi_2)=1$ $\therefore$ From equation (8), $R=\sqrt{A_1^2+A_2^2+2A_1A_2}=\sqrt{(A_1+A_2)^2}$ $\therefore R=\pm(A_1+A_2)$ b. When phase difference is $\dfrac{\pi}{2}$ radians: Two S.H.M.s are out of phase, i.e., $\phi_1-\phi_2=\dfrac{\pi}{2}$, $\therefore \cos(\phi_1-\phi_2)=0$ From equation (8), $R=\sqrt{A_1^2+A_2^2}$

3. 3.↑ back to the question

   Given: $x=\left[5\sin\pi t+12\sin\left(\pi t+\dfrac{\pi}{2}\right)\right]$ cm. This expression can be written as superposition of two parallel SHMs of same period, as $x=x_1+x_2$ where $x_1=5\sin\pi t=A_1\sin(\omega t+\phi_1)$ and $x_2=12\sin\left(\pi t+\dfrac{\pi}{2}\right)=A_2\sin(\omega t+\phi_2)$ $\therefore A_1=5$ cm, $A_2=12$ cm, $\omega=\pi$ rad/s $\phi_1=0$ and $\phi_2=\dfrac{\pi}{2}$ rad $\therefore$ Resultant amplitude, $A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos(\phi_1-\phi_2)}$ $=\sqrt{5^2+12^2+2\times5\times12\cos\left(-\dfrac{\pi}{2}\right)}$ $=\sqrt{169}=13$ cm Now, period, $T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\pi}=2$ rad/s Initial phase, $\delta=\tan^{-1}\left[\dfrac{A_1\sin\phi_1+A_2\sin\phi_2}{A_1\cos\phi_1+A_2\cos\phi_2}\right]$ $=\tan^{-1}\left[\dfrac{5\times\sin0+12\sin\left(\dfrac{\pi}{2}\right)}{5\times\cos0+12\cos\left(\dfrac{\pi}{2}\right)}\right]$ $=\tan^{-1}\left(\dfrac{12}{5}\right)=\tan^{-1}(2.4)=67°23'$