NSEP 2015 Problem 7 | Simple Harmonic Motion

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) $5$

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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