Simple Harmonic Motion
13.0 Vector Method of Combining Two or More Simple Harmonic Motions
13.0 Vector Method of Combining Two or More Simple Harmonic Motions
A simple harmonic motion is produced when a force (called restoring force) proportional to the displacement act on a particle. If a particle is acted upon by two such forces the resultant motion of the particle is a combination of two simple harmonic motions.
Suppose the two individual motions are represented by, $${x_1} = {A_1}sin\omega t\quad and\quad {A_2}{\text{sin}}\left( {\omega t + \phi } \right)$$
The resultant displacement of the particle is given by,
$$\begin{equation} \begin{aligned} x = {x_1} + {x_2} \\ x = {A_1}sin\omega t + {A_2}{\text{sin}}\left( {\omega t + \phi } \right) \\ x = {A_1}sin\omega t + {A_2}sin\omega tcos\phi + {A_2}cos\omega tsin\phi \\ x = \left( {{A_1} + {A_2}cos\phi } \right)sin\omega t + {A_2}sin\phi cos\omega t \\ \left( {{A_1} + {A_2}cos\phi } \right) = Acos\alpha \quad ..\left( i \right) \\ {A_2}sin\phi = Asin\alpha \quad ..\left( {ii} \right) \\\end{aligned} \end{equation} $$
Dividing equation $(ii)$ by $(i)$ we get,
$$tan\alpha = \frac{{{A_2}sin\phi }}{{{A_1} + {A_2}cos\phi }}$$
Squaring and adding equation $(i)$ & $(ii)$ we get,
$$\begin{equation} \begin{aligned} {A^2} = A_1^2 + A_2^2\left( {co{s^2}\phi + si{n^2}\phi } \right) + 2{A_1}{A_2}cos\phi \\ {A^2} = A_1^2 + A_2^2 + 2{A_1}{A_2}cos\phi \\ A = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}cos\phi } \\\end{aligned} \end{equation} $$
Example 9. A particle is subjected to two simple harmonic motions. $${x_1} = 4.0\sin \left( {100\pi t} \right)\quad and\quad {x_2} = 3.0sin\left( {100\pi t + \frac{\pi }{3}} \right)$$
Find:
(a) the equation of resultant SHM
(b) the displacement at $t=0$
(c) the maximum speed of the particle
(d) the maximum acceleration of the particle
Solution:
Given, $\phi = \frac{\pi }{3}$
(a). Amplitude of the resultant SHM,
$$\begin{equation} \begin{aligned} x = \sqrt {x_1^2 + x_2^2 + 2{x_1}{x_2}cos\phi } \\ x = \sqrt {{4^2} + {3^2} + 2 \times 4 \times 3 \times cos\frac{\pi }{3}} \\ x = \sqrt {16 + 9 + 12} = \sqrt {37} \\\end{aligned} \end{equation} $$
Angle made by resultant with SHM $x_1$ is $\alpha$,
$$\begin{equation} \begin{aligned} tan\alpha = \frac{{{x_2}sin\phi }}{{{x_1} + {x_2}cos\phi }} \\ tan\alpha = \frac{{3sin\frac{\pi }{3}}}{{4 + 3cos\frac{\pi }{3}}} = \frac{{3\sqrt 3 }}{{11}} \\ \alpha = ta{n^{ - 1}}\left( {\frac{{3\sqrt 3 }}{{11}}} \right) \\\end{aligned} \end{equation} $$
So, from Fig SHM 36, we get,
$$sin\alpha = \frac{{3\sqrt 3 }}{{2\sqrt {37} }}$$ Therefore the resultant SHM equation becomes,
$$x = \sqrt {37} {\text{sin}}\left( {100\pi t + \alpha } \right)\quad {\text{where }}\alpha = ta{n^{ - 1}}\left( {\frac{{3\sqrt 3 }}{{11}}} \right)$$
(b). At $'t'\ =\ 0$, $$\begin{equation} \begin{aligned} x = \sqrt {37} \sin \alpha = \sqrt {37} \times \frac{{3\sqrt 3 }}{{2\sqrt {37} }} \\ x = \frac{{3\sqrt 3 }}{2}{\text{units}} \\\end{aligned} \end{equation} $$
(c). Maximum speed ($v_max$) is $$\begin{equation} \begin{aligned} {v_{max}} = A\omega \\ {v_{max}} = \sqrt {37} \times 100\pi \\ {v_{max}} = 100\pi \sqrt {37} units \\\end{aligned} \end{equation} $$
(d). Maximum acceleration ($a_max$)
$$\begin{equation} \begin{aligned} {a_{max}} = A{\omega ^2} \\ {a_{max}} = \sqrt {37} \times {\left( {100\pi } \right)^2} \\ {a_{max}} = {\left( {100\pi } \right)^2}\sqrt {37} units \\\end{aligned} \end{equation} $$
