CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    The equation of a transverse wave travelling along positive \[x\] axis with amplitude 0.2 m, velocity 360 m/sec and wavelength 60 m can be written as:

    A)  \[y=0.2\sin \pi \left[ 6t+\frac{x}{60} \right]\]     

    B)  \[y=0.2\sin \pi \left[ 6t-\frac{x}{60} \right]\]

    C)  \[y=0.2\sin 2\pi \left[ 6t-\frac{x}{60} \right]\]

    D)  \[y=0.2\sin 2\pi \left[ 6t+\frac{x}{60} \right]\]

    Correct Answer: C

    Solution :

    Given \[a=0.2\] metre \[\upsilon =360\,m/s,\,\lambda =60\] metre Equations of transverse wave travelling along positive x-axis                 \[y=a\sin 2\pi \left[ \frac{t}{T}-\frac{x}{\lambda } \right]\] or            \[y=a\sin 2\pi \left[ \frac{\upsilon }{\lambda }t-\frac{x}{\lambda } \right]\]                 \[y=0.2\sin 2\pi \left[ 6t-\frac{x}{60} \right]\]


You need to login to perform this action.
You will be redirected in 3 sec spinner