Question

The displacement (in $\text{m}$) equation of a particle is $x=3\text{sin}2t+4\text{cos}2t$. The amplitude (in $\text{m}$) and maximum velocity (in $\text{m/s}$) will be respectively.

• A. $5,10$
• B. $3,2$
• C. $4,2$
• D. $3,4$

Answer 1

The correct option is A $5,10$

Given,
$x=3\text{sin}2t+4\text{cos}2t$
Rewriting the above equation as,
$x=3\text{sin}2t+4\text{sin}(2t+\frac{\pi }{2})....\left(i\right)$
$\because \sin (\frac{\pi }{2}+\theta )=\cos \theta$
Equation $\left(i\right)$ represents that displacement $\left(x\right)$ of particle is due to superposition of two independent $\text{SHM}$, with a phase difference of $\frac{\pi }{2}$
$\Rightarrow x=a_1\sin \omega t+a_2\sin (\omega t+\varphi )$

From the phases diagram given below:

The resultant amplitude of the particle will be given by,
$\Rightarrow a=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos \varphi }$
Where amplitude of individual $SHM$ is, $a_1=3\text{m},\&a_2=4\text{m}$ and phase difference $\varphi =\frac{\pi }{2}$
$\Rightarrow a=\sqrt{a_1^2+a_2^2}=\sqrt{3^2+4^2}$
$\therefore a=5\text{m}$
Maximum velocity of the particle is given by,
$\Rightarrow \left(v_{\text{max}}\right)=a\omega$
$\Rightarrow v_{\text{max}}=5\times 2=10\text{m/s}$