JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Waves

  • question_answer
    A transverse sinusoidal wave, moves along a string in the positive x-direction at a speed of 10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snap-shot of the wave is shown in figure. The velocity of point P when its displacement is 5 cm is  

    A)  \[\frac{\sqrt{3\pi }}{50}\hat{j}m/s\]

    B)  \[-\frac{\sqrt{3\pi }}{50}\hat{j}m/s\]

    C) \[\frac{\sqrt{3\pi }}{50}\hat{i}m/s\]

    D) \[-\frac{\sqrt{3\pi }}{50}\hat{i}m/s\]

    Correct Answer: A

    Solution :

    [a] Since the wave is sinusoidal moving in positive x-axis the point will move parallel to y-axis therefore options [c] and [d] are ruled out. As the wave moves forward in positive X-direction, the point should move upwards i.e. in the positive Y-direction. Therefore correct option is a. Alternate solution-1: Equation of a wave moving in positive x-axis is given as \[y=A\sin (\omega t-\phi )\] or \[{{v}_{p}}=A\omega \,\cos \,(\omega t-\phi )\] Here y= 5 cm, A= 10 cm, \[\therefore \,\,5=10\sin (\omega t-\phi )\Rightarrow \omega t-\phi =30{}^\circ \] Substituting this value in the equation of velocity we get \[{{v}_{p}}=0.10\times \omega \,\cos 30{}^\circ \] Now \[v=v\lambda \,\,\,\therefore \,\,\,v=\frac{v}{\lambda }=\frac{0.10}{0.5}=0.2\] \[\therefore \,\,\,\omega =2\pi v=2\pi \times 0.2=0.4\pi \] \[\Rightarrow \,{{v}_{p}}=0.1\times 0.4\pi \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{50}\pi \] It has to be in positive y direction.


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