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

  • question_answer
    The amplitude of a wave disturbance propagating in the positive x-direction is given by \[y=\frac{1}{1+{{x}^{2}}}\] at t=0 and \[y=\frac{1}{2+{{x}^{2}}-2x}\] at t = 2s, where x and y are in meter. Assuming that the shape of the wave disturbance does not change during the propagation, the speed of the wave is

    A)  \[0.5m/s\]         

    B)  \[~1m/s\]

    C) \[1.5m/s\]          

    D) \[2\text{ }m/s\]

    Correct Answer: A

    Solution :

    [a] At, \[t=0,\,y=\frac{1}{1+{{x}^{2}}}\,\] or \[x=\sqrt{\frac{1-y}{y}}={{x}_{1}}\] At \[t=2s,\]  \[y=\frac{1}{2+{{x}^{2}}-2x}=\frac{1}{1+{{(x-1)}^{2}}}\]  or  \[{{(x-1)}^{2}}=\frac{1-y}{y}\]  or \[x=1+\sqrt{\frac{1-y}{y}}={{x}_{2}}\] \[\therefore \,\,\,\,Speed\text{ }of\text{ }the\text{ }wave\] \[v=\frac{\Delta x}{\Delta t}=\frac{{{x}_{2}}-{{x}_{1}}}{{{t}_{2}}-{{t}_{1}}}=\frac{1}{2-0}=0.5m/s\]


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