Try out this problem on Simple Harmonic Motion (SHM) from National Standard Examination in Physics 2015-2016.
A particle execute a periodic motion according to the relation $x^2 = 4 \cos^2(50t) \sin^2(500t)$. Therefore, the motion can be considered to be the superposition of $n$ independent simple harmonic motions, where $n$ is
(a) $2$ (b) $3$ (c) $4$ (d) $5$
Basic Trigonometry
Equations of SHM
Concept of Physics H.C. Verma
University Physics by H. D. Young and R.A. Freedman
Fundamental of Physics D. Halliday, J. Walker and R. Resnick
National Standard Examination in Physics(NSEP) 2015-2016
Option-(b) $3$
For a simple harmonic motion the equation is $X$ = $A \cos(wt + \phi )$ or $X = A \sin(wt + \phi )$, where X is the position of the particle, A is the amplitude and w is the angular frequency and $\phi $ is the initial phase. We can also consider it to be zero.
If there are $n$ harmonic oscillator then the resultant can be written as ,
$X$=$A_{1} \sin \left(w_{1} t+\phi_{1}\right)$+$A_{2} \sin \left(w_{2} t+\phi_{2}\right)$+$\cdots$+$A_{n} \sin \left(w_{n} t+\phi_{n}\right)$
For this problem, we just have to use suitable trigonometric identities to convert them into sum.
$X$ =$ x^2$ = $4 \cos (50t) \sin (500t) \cos (50t)$, Now, we will use $2\cos(\theta)\sin(\phi)$= $\sin(\theta +\phi)$ +$ \sin(\theta -\phi)$, hence,
$X$=$2(\sin (550 t)$+$\sin (450 t)) \cos (50 t)$=$2 \sin (550 t) \cos (50 t)$+$2 \sin (450 t) \cos (50 t)$
Again applying the same identity,
$X$=$\sin (600 t)$+$2 \sin (500 t)$+$\sin (400 t)$
Hence, there are $3$ SHM.
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