2. Solved Papers
  3. Haryana PMT

Haryana PMT Haryana PMT Solved Paper-2003

  • question_answer
    A   transverse  wave   is  given  by \[y\] = A \[\sin 2\pi (ft-x/\lambda )\].The maximum particle velocity is 4 times the wave velocity when :

    A)  \[\lambda =2\pi A\]                      

    B)  \[\lambda =\pi A\]

    C) \[\lambda =\pi A/2\]                     

    D) \[\lambda =\pi A/4\]

    Correct Answer: C

    Solution :

                    Maximum particle velocity              \[{{\upsilon }_{m}}=A\omega =A\times 2\pi f\]                 (\[\because \] \[c=f\lambda ,\] \[\Rightarrow \] \[f=\frac{c}{\lambda }\])                 \[{{\upsilon }_{m}}=\frac{A2\pi c}{\lambda }\]           ?..(1) If \[{{\upsilon }_{m}}=4c\], then from equation (1)                            \[4c=\frac{A2\pi c}{\lambda }\] So,          \[\lambda =\frac{\pi A}{2}\]


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