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Question: A transverse sinusoidal wave is propagating in negative x-direction in a string stretched along x-ax...

A transverse sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis. A particle of string at x=2mx=2m is found at its mean position and it is moving in positive y-direction at t=1st=1s. The amplitude of the wave, the wavelength and the angular velocity of the wave are 0.1 m, π4m\dfrac{\pi }{4}m and 4π4\pi respectively. The equation of the wave is –

& \text{A) }y=0.1\sin (4\pi (t-1)+8(x-2)) \\\ & \text{B) }y=0.1\sin (4\pi (t-1)-8(x-2)) \\\ & \text{C) }y=0.1\sin (4\pi t+8x) \\\ & \text{D) }y=0.1\sin (4\pi t-8x) \\\ \end{aligned}$$
Explanation

Solution

We need to understand how a sinusoidal wave equation can be explained completely using an equation. The amplitude, the wavelength and the frequency of the wave can be combined in order to get the required solution for this problem.

Complete Step-by-Step Solution:
We are given the physical quantities such as the amplitude, the wavelength and the angular frequency of a sinusoidal wave. We know that these quantities are the elements which best describes a given wave and distinguishes it from the other waves.

We know that the general form of sinusoidal wave involves the amplitude, the propagation constant, the angular frequency and a phase shift if any. It is given as –
y=Asin(ωt±kx+ϕ)y=A\sin (\omega t\pm kx+\phi )
Where, y is the displacement at an instant t, A is the amplitude of the wave, ω\omega is the angular frequency, t is the time at any instant, k is the propagation constant, x is the direction of propagation and ϕ\phi is the phase constant.

The propagation constant can be obtained from the wavelength as –

& k=\dfrac{2\pi }{\lambda } \\\ & \Rightarrow k=\dfrac{2\pi }{\dfrac{\pi }{4}} \\\ & \therefore k=8 \\\ \end{aligned}$$ Now, we can obtain the required wave equation using the given quantities as – $$\begin{aligned} & \text{At, }t=t-1 \\\ & x=x-2 \\\ & \omega =4\pi \\\ & k=8 \\\ & \phi ={{0}^{0}} \\\ & A=0.1m \\\ & \Rightarrow y=A\sin (\omega t+kx+\phi ) \\\ & \therefore y=0.1\sin (4\pi (t-1)+8(x-2)) \\\ \end{aligned}$$ This is the required solution. **The correct answer is option A.** **Note:** The wave equation is the most convenient and easy way of displaying a wave. It includes all its essential quantities that is required to compare it with another wave. This equation is obtained very easily from the second order differentials which makes them even easier.