Explanation:-

here $\lambda =6m$

$\therefore k=\frac{2\pi }{6}=\frac{\pi }{3}$

$\omega =\frac{2\pi }{T}=\frac{2\pi }{0.5}=4\pi$

given,amplitude =5m

$y=5sin(\frac{\pi }{3}x-4\pi t)$

[the standard equation is y=Asin(kx-$\omega$ t)]

Explanation:-

$y=Asi{n}^2(kx-\omega t)$

This  is not ideal equation of wave we don't have$si{n}^2\theta$ in ideal equation so let's remove the square.

Use trigonometry

$cos2\theta =co{s}^2\theta -si{n}^2\theta$

$=1-2si{n}^2\theta$

$si{n}^2\theta =\frac{1-cos2\theta }{2}$

$y=A\frac{\{1-cos2(kx-wt{)}^2\}}{2}$

$y=\frac{A}{2}-\frac{A}{2}cos(2kx-2\omega t)$

$y^{\prime }=(y-\frac{A}{2})=-\frac{A}{2}cos(2kx-2\omega t)$

So amplitude =$-\frac{A}{2}$

angular frequency$\left({\omega }^{\prime }\right)=2\omega$

frequency(f)=$\frac{\omega }{2\pi }=\frac{2\omega }{2\pi }=\frac{\omega }{\pi }$

Explanation:-

From the y-t graph we can see the wave repeats itself on interval of 4s. So the time period is 4s. from the y-x graph we see that the distance between the points after which the wave repeats itself is 4m. Hence the wave length is 4m.

$f=\frac{1}{T}=\frac{1}{4}{S}^{-1}=0.25s^{-1}$

$v=\lambda f=4\frac{1}{4}=1m{s}^{-1}$

$v_{p}max=A\omega$

$=A\times (2\pi f$

$=3\times \times 2\pi \times 0.25$  [A=3 from y vst graph]

$a_{p}max={\omega }^{2}A$

$=(2\pi \times 0.25{)}^2\times 3$

$\frac{3}{4}{\pi }^2=0.075{\pi }^{2}mm{s}^{-2}$

Explanation:-

Given shape of the wave at t=0

$g\left(x\right)=Asin\left(\frac{x}{a}\right)$

let's say the wave looks like

It's wave speed is v. After time t, the wave would have moved vt.

Now if at t = 0

The equation is $g\left(x\right)=Asin\left(\frac{x}{a}\right)$

This wave was at x = x, t= t will be same as the wave at x= x-vt at t = 0 

So at t = 0 g(x) = A sin $\frac{x-vt}{a}$

Alternate Solution:- Given g(x) = A sin $\left(\frac{x}{a}\right)$

General equation A sin (kx - $\omega$t)

Explanation:-

Here given is $g(x,t_0)=Asin\left(\frac{x}{a}\right)$  at $t=t_0$

Comparing with the general equation g(x,t)=$Asin(kx-\omega t+\varphi )$

We get at $t=t_0$

$k=\frac{1}{a}$

also$-\omega t_0+\varphi =0$

$\Rightarrow \varphi =\omega t_0$

We also known that $k=\frac{\omega }{v}$

$\Rightarrow \omega =\frac{v}{a}$

so we get

$g(x,t)=Asin(\frac{x}{a}-\frac{vt}{a}+\frac{v{t}_0}{a})$

$g(x,t)=Asin[\frac{x}{a}-\frac{vt}{a}+\frac{v{t}_0}{a}]$

$=Asin\left(\frac{x-v(t-t_0)}{a}\right)$

Explanation:-

The standard wave equation for a wave is $y=Asin(kx-\omega t)$

Given equation $y=3sin6.28(0.5x-50t)$

$\Rightarrow y=3sin(6.28\times 0.5x-6.28\times 50t)$

Comparing we get

amplitude A = 3 cm

Right here, we can see that B is the only possible option, but we may as well calculate the other values too.

Wave number $k=\frac{2\pi }{\lambda }=6.28\times 0.5$

$\Rightarrow \lambda =\frac{2\pi }{k}=2cm$

Angular frequency $\omega =6.28\times 50$

Frequency $f=\frac{1}{T}=\frac{\omega }{2\pi }=\frac{6.28\times 50}{2\times 3.14}$

$f=50Hz$

wave speed =$\frac{\lambda }{T}=2\times 50=100\frac{cm}{s}$

Explanation:-

crest to crest distance is wavelength λ = 800 km 

Time period: Time taken for wave to travel a distance of one wavelength = 1 hr

Wave velocity = $\frac{Distance}{Time}=\frac{1wavelength\left(\lambda \right)}{Timeperiod\left(T\right)}$

$\frac{800km}{1}=800km/hr$

Explanation:-

$y=A{e}^{-{(\frac{x}{a}+\frac{t}{T})}^2}$

at t=T the pulse would be

$y=A{e}^{-{(\frac{x}{a}+1)}^2}$

$\frac{dy}{dx}=\frac{-2A}{a}(\frac{x}{a}+1)e^{-{(\frac{x}{a}+1)}^2}$

$\frac{dy}{dx}=0\Rightarrow \frac{x}{a}+1=0$

$x=-a$

Explanation:-

According to principle of super position

${\vec{y}}_{net}=\overrightarrow{y_1}+\overrightarrow{y_2}$

Here $y_1\&y_2$ are particle displacement vectors

differentiating with respect to time.

$\frac{d{\vec{y}}_{net}}{dt}=\frac{d\overrightarrow{y_1}}{dt}+\frac{d\overrightarrow{y_2}}{dt}$

${\vec{v}}_{net}={\vec{v}}_1+{\vec{v}}_2$

Here velocities are vectors.

So yes the superposition law is applicable to velocities.

Now let us square both sides

$(v_{net}{)}^2=({\vec{v}}_1+{\vec{v}}_2{)}^2$

$v_{net}^2={v_1}^2+{v_2}^2+2v_1{v}_2$ (all are magnitudes)
 

So clearly the law of super position is not applicable for net kinetic energy.

Explanation:-