Correct Answer is Option (B)

We can write the given equation as:

$y=\sqrt{(\sin \omega t{)}^2+(\cos \omega t{)}^2}\cos (\omega t-\arctan \frac{\sin \omega t}{\cos \omega t})$

Using the identity ${\sin }^2\theta +{\cos }^2\theta =1$, we get:

$y=\sqrt{1+\sin 2\omega t}\cos (\omega t-\frac{\pi }{4})$

The amplitude of the motion is the maximum value of $|y|$, which occurs when $\sin 2\omega t=1$, i.e., at $t=\frac{\pi }{4\omega }+\frac{n\pi }{\omega }$, where $n$ is an integer. Substituting this value of $t$ in the above equation, we get:

$|y_{\text{max}}|=\sqrt{1+\sin \frac{\pi }{2}}=\sqrt{2}$

Therefore, the amplitude of the motion is $\sqrt{2}$

Correct Answer is 20

Given, $y_1=10\sin (\omega t+\frac{\pi }{3})\mathrm{cm}$ and

$y_2=5(\sin \omega t+\sqrt{3}\cos \omega t)$

$=10\sin (\omega t+\frac{\pi }{3})$

Thus the phase difference between the waves is 0 .

$\text{ so }A=A_1+A_2=20\mathrm{cm}$

Correct Answer is 120

$A_R=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$

$\begin{array}{rl}& 8=\sqrt{8^2+8^2+2\times 8\times 8\cos \varphi }\\ \\ & \Rightarrow \cos \varphi =-\frac{1}{2}\\ \\ & \Rightarrow \varphi ={120}^{\circ }\end{array}$

Correct Answer is Option (A)

$y_1=5\sin (2\pi x-2\pi vt)$

$y_2=3\sin (2\pi x-2\pi vt+3\pi )$

$\Rightarrow$ Phase difference = 3$\pi$

$\Rightarrow A_{net}=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos (3\pi )}$

$\Rightarrow A_{net}=2$ cm

Correct Answer is 60

$A_{net}=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$

$\sqrt{3}A=\sqrt{A^2+A^2+2A^2\cos \varphi }$

$3A^2=2A^2+2A^2\cos \varphi$

$\cos \varphi =\frac{1}{2}$

$\varphi ={60}^{\circ }$