A transverse wave of frequency 500 ,Hz and sp | vedclass

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Question

$(b)$ Equation for a travelling wave is

$y= a \sin \left(\omega t-k x+\phi_{0}\right)$

As at $t=0$ at $x=0$ and $y=0$,

$\Rightarrow \quad \ph

Vedclass · Accepted Answer

$-0.02$

Vedclass · Answer

$(b)$ Equation for a travelling wave is

$y= a \sin \left(\omega t-k x+\phi_{0}ight)$

As at $t=0$ at $x=0$ and $y=0$,

$\Rightarrow \quad \phi_{0}=0$

So, equation of wave over string is

$y= a \sin (\omega t-k x)$

Here,

$\omega=2 \pi f=2 \pi 	imes 500$

$=1000 \pi\left(\frac{ rad }{ s }ight)$

Angular wave number,

$k=\frac{\omega}{v}=\frac{1000 \pi}{100}=10 \pi\left( m ^{-1}ight)$

So, wave equation is

$y=a \sin (1000 \pi \cdot t-10 \pi \cdot x)$

Given at $t=0$ at $x=0.25 m$ and $y=0.02 m$.

So, $\quad 0.02=a \sin (-10 \pi 	imes 0.25)$

$=-a \sin \left(\frac{5}{2} \piight)$

$[	herefore \sin (-	heta)=-\sin 	heta]$

$\Rightarrow \quad 0.02=-a \sin \left(2 \pi+\frac{\pi}{2}ight)$

$\Rightarrow \quad 0.02=-a \sin \left(\frac{\pi}{2}ight)$ or $\alpha =-0.02$

Here note that $a$ is amplitude and its positive and negative values are same. When we are getting a negative value this means particle is displaced below mean position.

So, we have

$y=-0.02 \sin (1000 \pi t-10 \pi x)$

Now, at $t=5 	imes 10^{-4} s$ and $x=0.2 m$, value of displacement of particle is

$y =-0.02 \sin \left(1000 \pi 	imes 5 	imes 10^{-4}ight.$

$=-0.02 \sin \left(\frac{\pi}{2}-2 \piight)$

$=0.02 \sin \left(2 \pi-\frac{\pi}{2}ight)$

${\left[	herefore \sin \left(\frac{\pi}{2}-2 \piight)=-\sin \left(2 \pi-\frac{\pi}{2}ight)ight.}$

$\quad 	ext { Also, } \sin (2 \pi-	heta)=-\sin 	heta]$

$=-0.02\,m$

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