JEE Main & Advanced Sample Paper JEE Main Sample Paper-20

  • question_answer
    DIRECTION (Qs. 53): Read the following passage and answer the questions that follows:
    A sinusoidal wave is propagating in negative x-direction in a string stretched along x-axis. A particle of string at \[x=2m\] is found at its mean position and it is moving in positive y-direction at \[t=1\] sec. The amplitude of the wave, the wavelength and the angular frequency of the wave are \[0.1\] meter, \[\pi /2\] meter and \[2\pi \] rad/sec respectively.
    The speed of particle at \[x=2m\] and \[t=1\sec \] is

    A)  \[0.271\text{ }m/s\]                  

    B)  \[0.671\text{ }m/s\]

    C)  \[0.471\text{ }m/s\]                   

    D)    0

    Correct Answer: A

    Solution :

     As given the particle at \[x=2\] is at mean position at \[t=1\] sec. \[\therefore \]  Its velocity \[v=\omega A=2\pi \times 0.1=0.2\pi \,m/s\].


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