Answer
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Hint: A wave traveling in the positive direction along x from one point to another point of a medium is known as a progressive wave or a traveling wave. When a progressive wave travels in a medium, then the medium's particles vibrate in the same way. Still, the vibration phase changes from one particle to particle at any instant.
If $\phi $ be the initial phase angle, then the progressive wave traveling along the positive x-axis direction is represented as:
${\rm{y}}\; = \;{\rm{a sin(kx - }}\omega {\rm{t + }}\phi {\rm{)}}$
Complete step by step answer:
Now according to the question:
The amplitude of the wave is given as ${\rm{A}}\;{\rm{ = }}\;{\rm{2}}{\rm{.0cm = 0}}{\rm{.02m}}$
Wavelength is given as $\lambda \; = \,1{\rm{m}}$
Now the wave number $k\; = \;\dfrac{{2\pi }}{\lambda }\, = \;2\pi {{\rm{m}}^{ - 1}}$
And the angular frequency $\omega \; = \;vk\; = \;5m/s \times 2\pi {\rm{m}}{{\rm{s}}^{ - 1}}\; = 10\pi {\rm{rad}}{{\rm{s}}^{ - 1}}$
Therefore, we put the values in the above equation with coordinates x and t:
$ \Rightarrow {\rm{y(x,t) = (0}}{\rm{.02) sin}}\left[ {2\pi (x - 5.0t) + \phi } \right]$
We have given that for x=0 and t=0,
${\rm{y}}\; = \;0\;{\rm{and}}\;\dfrac{{\delta y}}{{\delta x}} < 0$
$ \Rightarrow \; - 0.02{\rm{sin}}\phi \; = 0\,({\rm{as y = 0)}}$
$\therefore \; - 0.2\pi \cos \phi < 0$
From these conditions, we include that,
$\phi \; = \;2{\rm{n}}\pi {\text{ where n = 0,2,4,6}}......$
Therefore, ${\rm{y}}\;{\rm{ = (0}}{\rm{.02m) }}\left[ {\sin (2\pi {{\rm{m}}^{ - 1}})x - (10\pi {{\rm{s}}^{ - 1}}){\rm{t}}} \right]$
Hence, the correct option is (C).
Note:
In SI, the unit of propagation constant or angular wave number (k) radian /meter. The Dimensional formula for the angular wave number is $\left[ {{{\rm{M}}^0}{{\rm{L}}^{ - 1}}{{\rm{T}}^0}} \right]$ . Also, progressive wave traveling along the positive x-axis with a speed $v$ can be represented as ${\rm{y = a sin}}\dfrac{{2\pi }}{\lambda }\;(vt - x)$
If $\phi $ be the initial phase angle, then the progressive wave traveling along the positive x-axis direction is represented as:
${\rm{y}}\; = \;{\rm{a sin(kx - }}\omega {\rm{t + }}\phi {\rm{)}}$
Complete step by step answer:
Now according to the question:
The amplitude of the wave is given as ${\rm{A}}\;{\rm{ = }}\;{\rm{2}}{\rm{.0cm = 0}}{\rm{.02m}}$
Wavelength is given as $\lambda \; = \,1{\rm{m}}$
Now the wave number $k\; = \;\dfrac{{2\pi }}{\lambda }\, = \;2\pi {{\rm{m}}^{ - 1}}$
And the angular frequency $\omega \; = \;vk\; = \;5m/s \times 2\pi {\rm{m}}{{\rm{s}}^{ - 1}}\; = 10\pi {\rm{rad}}{{\rm{s}}^{ - 1}}$
Therefore, we put the values in the above equation with coordinates x and t:
$ \Rightarrow {\rm{y(x,t) = (0}}{\rm{.02) sin}}\left[ {2\pi (x - 5.0t) + \phi } \right]$
We have given that for x=0 and t=0,
${\rm{y}}\; = \;0\;{\rm{and}}\;\dfrac{{\delta y}}{{\delta x}} < 0$
$ \Rightarrow \; - 0.02{\rm{sin}}\phi \; = 0\,({\rm{as y = 0)}}$
$\therefore \; - 0.2\pi \cos \phi < 0$
From these conditions, we include that,
$\phi \; = \;2{\rm{n}}\pi {\text{ where n = 0,2,4,6}}......$
Therefore, ${\rm{y}}\;{\rm{ = (0}}{\rm{.02m) }}\left[ {\sin (2\pi {{\rm{m}}^{ - 1}})x - (10\pi {{\rm{s}}^{ - 1}}){\rm{t}}} \right]$
Hence, the correct option is (C).
Note:
In SI, the unit of propagation constant or angular wave number (k) radian /meter. The Dimensional formula for the angular wave number is $\left[ {{{\rm{M}}^0}{{\rm{L}}^{ - 1}}{{\rm{T}}^0}} \right]$ . Also, progressive wave traveling along the positive x-axis with a speed $v$ can be represented as ${\rm{y = a sin}}\dfrac{{2\pi }}{\lambda }\;(vt - x)$
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