BVP Medical BVP Medical Solved Paper-2007

  • question_answer
    A man is watching two trains, one leaving and the other coming with equal speed of 4 m/s. If they sound their whistles each of frequency 240 Hz, the number of beats heard by man (velocity of sound in air = 320 m/s) will be equal to

    A)                 12                                          

    B)                 zero

    C)                 3                                             

    D)                  6

    Correct Answer: D

    Solution :

                    Key Idea: The number of beats heard by man is the difference of apparent frequencies of two trains. From Dopplers effect, the perceived frequency, when train is approaching the man is given by \[\Rightarrow \] where v is speed of sound, \[{{n}_{2}}=\frac{50}{49}\times 392=400\] is speed of source. Given, \[={{n}_{2}}-{{n}_{1}}=\]                 \[=400-392=8\]                 \[y=2\,a\,\sin \,\frac{2\pi }{\lambda }\,x\cos \frac{2\pi }{\lambda }ct\] When train is receding the perceive ding frequency is                 \[\lambda \]                 \[y=5\sin \frac{\pi \,\,x}{3}\,\cos \,40\,\pi \,t\]                 \[y=5\sin \frac{\pi \,\,x}{3}\,\cos \,40\,\pi \,t\] Number of beats per second (beat frequency) \[m/{{s}^{2}}\]difference   of   the   apparent frequencies of sound sources. \[y=a\cos (\omega t-kx),\]          \[\left[ {{M}^{0}}{{L}^{-1}}{{T}^{-1}} \right]\]


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