JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    The displacement y of a particle in a medium can be expressed as\[y={{10}^{-6}}\sin \left( 100t+20x+\frac{\pi }{4} \right)m\], where t is in second and \[x\]in metre. The speed of the wave is

    A) \[2000\,m/s\]   

    B)        \[5\,\,m/s\]       

    C)        \[20\,\,m/s\]    

    D)        \[5\pi \,\,m/s\]

    Correct Answer: B

    Solution :

    As given \[y={{10}^{-6}}\sin \left( 100t+20x+\frac{\pi }{4} \right)\]                   ...(i) Comparing it with \[y=a\sin (\omega t+kx+\phi )\]                                            ?(ii) We find,\[\omega =100\,rad/s,k=20/m\] \[\therefore \,\,\,\,\,\,\,v=\frac{\omega }{k}=\frac{100}{20}=\,5m/s\]


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