Question

The displacement of a particle executing simple harmonic motion is given by
$x\left(t\right)=2\sin \left(2\pi t\right)+2\cos (2\pi t+\frac{\pi }{6})$
The total energy of the particle is $(m=1\text{kg}$)

• A. $2{\pi }^2\text{J}$
• B. $8{\pi }^2\text{J}$
• C. $6{\pi }^2\text{J}$
• D. $3{\pi }^2\text{J}$

Answer 1

The correct option is B $8{\pi }^2\text{J}$

Given that
$x\left(t\right)=2\sin \left(2\pi t\right)+2\cos (2\pi t+\frac{\pi }{6})$

$x\left(t\right)=2\sin \left(2\pi t\right)+2\cos 2\pi t\cos \frac{\pi }{6}-2\sin 2\pi t\sin \frac{\pi }{6}$

$x\left(t\right)=(2-2\sin \frac{\pi }{6})\sin 2\pi t+2\cos \frac{\pi }{6}\cos 2\pi t...\left(1\right)$

Normal equation of $\text{SHM}$ is given by

$x\left(t\right)=A\sin (\omega t+\varphi )$

$x\left(t\right)=A\sin \omega t\cos \varphi +A\cos \omega t\sin \varphi .....\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$,

Here, $\omega =2\pi$ and
$A\cos \varphi =2-2\sin \frac{\pi }{6}$
$A\sin \varphi =2\cos \frac{\pi }{6}$

$\therefore A=\sqrt{(2-2\sin \frac{\pi }{6}{)}^2+(2\cos \frac{\pi }{6}{)}^2}=2$

Total Energy of the particle executing $\text{SHM}$ is given by

$E=\frac{1}{2}K{A}^2=\frac{1}{2}m{\omega }^2{A}^2=8{\pi }^2\text{J}$

Hence, option $\left(\text{B}\right)$ is correct.

Why this question? When two or more SHM occurs simultaneously, try to equate equation of single SHM.