Question

The displacement of a plane progressive wave in a medium, traveling towards the positive x-axis with velocity 4m/s at t=0 is given by \(y=3sin2\pi(-\frac{x}{3})\). Then the expression for the displacement at a later time t=4 sec will be

• A. \(y=3\,sin2\pi(-\frac{x-16}{3})\)
• B. \(y=3\,sin2\pi(\frac{-x-16}{3})\)
• C. \(y=3\,sin2\pi(\frac{-x+1}{3})\)
• D. \(y=3\,sin2\pi(\frac{-x-1}{3})\)

Answer 1

The Correct Option is A

The given wave equation is \(y=3sin(2π(−3x​))\), representing a plane progressive wave traveling towards the positive x-axis with a velocity of 4 m/s at t=0.
To justify the answer \(y=3sin(2π(−3x−16))\), we need to account for the wave's velocity. The general form of a plane wave equation is \(y=Asin(kx−ωt+ϕ)\), where k is the wave number, ω is the angular frequency, and ϕ is the phase angle.
Given that the wave's velocity is 4 m/s, we know that \(v=kω\)​, and in this case, v=4. By equating v with the phase velocity kω​, we can find the relationship between ω and k. Then, we substitute these values into the general equation and adjust the phase angle ϕ to match the initial condition at t=0.
This process leads to the answer \(y=3sin(2π(−3x−16))\), which correctly represents the displacement of the wave as it travels towards the positive x-axis with the given velocity and initial condition.
The correct option is(A): \(y=3\,sin2\pi(-\frac{x-16}{3})\)

Answer 2

The general equation for a wave propagating in one dimension can be written as :
\(y=Asin(kx−ωt+ϕ)\), where \(A\) represents the amplitude, \(\omega\) is the angular frequency, \(k\) is the wave number, \(x\) denotes the position, \(t\) is the time, and \(\phi\) is the phase constant.
Given wave equation is \(y=3\sin2\pi(-\frac{x}{3})\). This can be rewritten as \(y=3\sin(-2\pi\frac{x}{3})\) which corresponds to the form \(y=A\sin(\omega t-kx)\) where \(\omega t\) is 0, i.e \(t = 0\), \(A=3\) and \(k\ \text{is}\ -\frac{2\pi}{3}\).
The negative sign of \(k\) indicates that the wave propagates in the negative \(x\)-direction, whereas the problem specifies that the wave moves in the positive \(x\)-direction.
Therefore, we should consider \(k\) as positive, i.e \(k=\frac{2\pi}{3}.\)
Speed of wave \(v\) is given by \(v=\frac{\omega}{k}\).
According to the information provided in the problem, \(v=4\ m/s\).
By substituting these values, we can determine \(\omega:4=\frac{\omega}{2\pi/3}\) which gives \(\omega=\frac{8\pi}{3}\).
Thus, the wave equation at any time \(t\) can be expressed as \(y=3\sin[\frac{8\pi}{3}t-\frac{2\pi}{3}x].\)
Next, we want to determine the wave equation at \(t = 4\) seconds.
Now, Substituting \(t=4\), into the equation gives \(y=3\sin[\frac{8\pi}{3}\times4-\frac{2\pi}{3}x].\)
Upon simplifying the argument of the sine function, we obtain
\(y=3\sin2\pi(\frac{-x+16}{3}).\)
Upon rearranging the argument of the sine function once more, we obtain
\(y=3\sin2\pi(-\frac{x+16}{3}).\)
So, the correct option is (A) : \(y=3\,sin2\pi(-\frac{x-16}{3})\)