⇒Superposition of Waves

Correct Option is (D)

$\begin{array}{rl}& \text{ Given: }y=6\sin (12\pi t-0.02\pi x+\frac{\pi }{3})\\ & \omega t=12\pi t\\ & \Rightarrow \frac{2\pi }{T}=12\pi \end{array}$

$\begin{array}{rl}\therefore T& =\frac{1}{6}\\ \Rightarrow n& =6\end{array}$

Also given $\frac{2\pi }{\lambda }=0.02\pi$

$\therefore \lambda =\frac{2}{0.02}=100$

From equation: $\mathrm{v}=\mathrm{n}\lambda$, we get

$\begin{array}{rl}v& =6\times 100\\ & =600\mathrm{m}/\mathrm{s}\end{array}$

Correct Option is (B)

General equation of a simple harmonic progressive wave,

$y=A\sin 2\pi [\frac{t}{T}-\frac{x}{\lambda }]$

Given: $\mathrm{y}=\mathrm{a}\sin 2\pi (bt-cx)$

$\Rightarrow \mathrm{A}=\mathrm{a},\frac{1}{\mathrm{T}}=\mathrm{b}\text{ and }\frac{1}{\lambda }=\mathrm{cx}$

Also, ${\left({\mathrm{v}}_{\mathrm{p}}\right)}_{max}=\mathrm{a}\omega =\mathrm{a}(2\pi \mathrm{n})=\frac{\mathrm{a}2\pi }{\mathrm{T}}$

$=\frac{\mathrm{A}2\pi }{\mathrm{T}}=\mathrm{a}2\pi \mathrm{b}$

From $\mathrm{v}=\frac{\lambda }{\mathrm{T}}=\frac{1/\mathrm{c}}{1/\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{c}}$

Given: ${\left({\mathrm{v}}_{\mathrm{p}}\right)}_{max}=\frac{1}{2}\mathrm{v}$

$\begin{array}{rl}& \Rightarrow 2\pi ab=\frac{1}{2}\times \frac{b}{c}\\ \therefore & c=\frac{1}{4\pi a}\end{array}$

Correct Option is (D)

$y=\frac{1}{\sqrt{a}}\sin \omega t\pm \frac{1}{\sqrt{b}}\sin (\omega t+\frac{\pi }{2})$

Here phase difference $=\frac{\pi }{2}$

The resultant amplitude

$=\sqrt{{\left(\frac{1}{\sqrt{a}}\right)}^2+{\left(\frac{1}{\sqrt{b}}\right)}^2}=\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a+b}{ab}}$

Correct Option is (D)

The two equations can be written as

${\mathrm{Y}}_1=2\sin 2\pi (\frac{4\mathrm{t}}{0.2}-\frac{4\mathrm{x}}{2})=2\sin 2\pi (\frac{\mathrm{t}}{0.05}-\frac{\mathrm{x}}{0.5})$

and ${\mathrm{Y}}_2=4\sin 2\pi (\frac{4\mathrm{t}}{0.16}-\frac{4\mathrm{x}}{1.6})=2\sin 2\pi (\frac{\mathrm{t}}{0.04}-\frac{\mathrm{x}}{0.4})$

Comparing with standard equation

$\mathrm{Y}=\mathrm{A}\sin 2\pi (\frac{\mathrm{t}}{\mathrm{T}}-\frac{\mathrm{x}}{\lambda })$

We get, for the first wave,

$\mathrm{T}=0.05\mathrm{s}\text{ and }\lambda =0.5\mathrm{m}$

For the second wave,

$\mathrm{T}=0.04\mathrm{s}\text{ and }\lambda =0.4\mathrm{m}$

Hence their periods (hence frequencies) are not same. Their wavelength are also not same.

For first wave velocity $=\frac{\lambda }{\mathrm{T}}=\frac{0.5}{0.05}=10\mathrm{m}/\mathrm{s}$

For second wave velocity $=\frac{0.4}{0.04}=10\mathrm{m}/\mathrm{s}$

Hence velocity in same,

Correct Option is (C)

$\mathrm{n}=50\mathrm{Hz},\mathrm{t}=0.01\mathrm{s}$

If $\mathrm{T}$ is the period and phase difference is $\varphi$

$\begin{array}{rl}& \text{ then, }\frac{\mathrm{t}}{\mathrm{T}}=\frac{\varphi }{2\pi }\\ & \therefore \varphi =2\pi \frac{\mathrm{t}}{\mathrm{T}}=2\pi \mathrm{nt}\\ & \therefore \varphi =2\pi \times 50\times 0.01=\pi \mathrm{rad}\end{array}$