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Question: The maximum transverse velocity and maximum transverse acceleration of a harmonic wave in one-dimens...

The maximum transverse velocity and maximum transverse acceleration of a harmonic wave in one-dimensional string are 1ms11ms^{-1} and 1ms21ms^{-2} respectively. The phase velocity of the wave is 1ms11ms^{-1} . The waveform is
A. sin(xt)\sin(x-t)
B.sin(2xt)\sin(2x-t)
C. sin(x2t)\sin (x-2t)
D. sin(x2t)\sin\left(\dfrac{x}{2}-t\right)
E.sin(xt2)\sin \left(x-\dfrac{t}{2}\right)

Explanation

Solution

To find the wave equation, we have first found the value for the various parameters which are involved in the wave equation , such as the amplitude , propagation constant angular velocity and the phase difference of the wave.
Formula used:
y=Asin(kxωt+ϕ)y=A \sin(kx-\omega t+\phi)

Complete step-by-step solution:
A wave is a propagation of disturbance, which results in the transfer of energy from one particle of the medium to the other. These undergo harmonic oscillation and are also called the sine waves. There are broadly two types of waves, namely the travelling wave and the standing wave. The standing waves are caused due to superimposition of two or more travelling waves.
We know that the wave equation is given as y=Asin(kxωt+ϕ)y=A \sin(kx-\omega t+\phi) where, yy is the displacement of the wave due to amplitude AA , kk is the propagation constant at a distance xx from the origin , ω\omega is the angular velocity at time tt and ϕ\phi is the phase difference of the wave.
To begin with, let us consider that the phase difference is ϕ=0\phi=0 with respect to the origin.
Then, we also know that vmax=ωAv_{max}=\omega A and amax=ω2Aa_{max}=\omega^2 A where, vv and aa is the velocity and acceleration of the wave.
Then, we have vmaxamax=ωAω2A\dfrac{v_{max}}{a_{max}}=\dfrac{\omega A}{\omega^2 A}
Since, vmax=1ms1v_{max}=1ms^{-1} and amax=1ms2a_{max}=1ms^{-2}
    ω=1rads1\implies \omega=1 rads^{-1}
Then, the velocity v=ωkv=\dfrac{\omega}{k}
    k=ωv=1rad  m1\implies k=\dfrac{\omega}{v}=1 rad\; m^{-1}
Similarly, amplitude A=vω=1mA=\dfrac{v}{\omega}=1m
Now substituting the values in the wave equation, we have
y=1sin(1x1t+0)y=1\sin(1x-1t+0)
Thus, the correct answer is option A. sin(xt)\sin(x-t)

Note: The wave equation can either be described as a sine wave, as shown above or can also be described as a cosine wave. In this question, we have used almost all the formulas related to a wave equation, hence to solve this sum, one must know the relation between the various parameters of the wave equation.