Question

The displacement of particle in a periodic motion is given by $y=4{\text{cos}}^2\left(\frac{t}{2}\right)\text{sin}1000t$. This displacement may be considered as the result superposition of $n$ independent harmonic oscillations. Here the value of $n$ is :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Given, $y=4{\cos }^2\left(\frac{t}{2}\right)\sin \left(1000t\right)$
using , ${\cos }^2\left(\frac{\theta }{2}\right)=\frac{1+\cos \theta }{2}$
we get
$y=2(1+\cos t)\sin \left(1000t\right)$
$y=2\sin \left(1000t\right)+2\cos t\sin \left(1000t\right)$
Further, using $\sin (A+B)-\sin (A-B)=2\cos A\sin B$

$\Rightarrow y=2\sin \left(1000t\right)+\sin \left(1001t\right)+\sin \left(999t\right)$
Above equation suggests that the displacement of particle $\left(y\right)$ is superpositon of $3$ independent simple harmonic oscillations.

Hence, option $\left(c\right)$ is correct.