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Question: A sinusoidal wave traveling in the positive direction of x on a stretched membrane string has amplit...

A sinusoidal wave traveling in the positive direction of x on a stretched membrane string has amplitude 2.0 cm, Wavelength 1m, and wave velocity 5.0m/s. At x=0 and t=0. It is given that displacement y=0 and yx<0\dfrac{{\partial y}}{{\partial x}} < 0. Express the wave function correctly in the form y  =f(x,t){\rm{y}}\;{\rm{ = f(x,t)}}:-
A. y  =(0.04m)[sin(πm1)x(10πs1)t]{\rm{y}}\;{\rm{ = (0}}{\rm{.04m) }}\left[ {\sin (\pi {{\rm{m}}^{ - 1}})x - (10\pi {{\rm{s}}^{ - 1}}){\rm{t}}} \right]
B. y  =(0.02m)cos2π(x5t){\rm{y}}\;{\rm{ = (0}}{\rm{.02m) cos 2}}\pi (x - 5t)
C. y  =(0.02m)[sin(2πm1)x(10πs1)t]{\rm{y}}\;{\rm{ = (0}}{\rm{.02m) }}\left[ {\sin (2\pi {{\rm{m}}^{ - 1}})x - (10\pi {{\rm{s}}^{ - 1}}){\rm{t}}} \right]
D. y  =(0.02m)cosπ(x5t+14){\rm{y}}\;{\rm{ = (0}}{\rm{.02m) cos}}\pi (x - 5t + \dfrac{1}{4})

Explanation

Solution

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 ϕ\phi be the initial phase angle, then the progressive wave traveling along the positive x-axis direction is represented as:
y  =  asin(kxωt+ϕ){\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 A  =  2.0cm=0.02m{\rm{A}}\;{\rm{ = }}\;{\rm{2}}{\rm{.0cm = 0}}{\rm{.02m}}
Wavelength is given as λ  =1m\lambda \; = \,1{\rm{m}}
Now the wave number k  =  2πλ=  2πm1k\; = \;\dfrac{{2\pi }}{\lambda }\, = \;2\pi {{\rm{m}}^{ - 1}}
And the angular frequency ω  =  vk  =  5m/s×2πms1  =10πrads1\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:
y(x,t)=(0.02)sin[2π(x5.0t)+ϕ]\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,
y  =  0  and  δyδx<0{\rm{y}}\; = \;0\;{\rm{and}}\;\dfrac{{\delta y}}{{\delta x}} < 0
  0.02sinϕ  =0(asy=0)\Rightarrow \; - 0.02{\rm{sin}}\phi \; = 0\,({\rm{as y = 0)}}
  0.2πcosϕ<0\therefore \; - 0.2\pi \cos \phi < 0
From these conditions, we include that,
ϕ  =  2nπ where n = 0,2,4,6......\phi \; = \;2{\rm{n}}\pi {\text{ where n = 0,2,4,6}}......
Therefore, y  =(0.02m)[sin(2πm1)x(10πs1)t]{\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 [M0L1T0]\left[ {{{\rm{M}}^0}{{\rm{L}}^{ - 1}}{{\rm{T}}^0}} \right] . Also, progressive wave traveling along the positive x-axis with a speed vv can be represented as y=asin2πλ  (vtx){\rm{y = a sin}}\dfrac{{2\pi }}{\lambda }\;(vt - x)