2. Solved Papers
  3. JCECE Engineering

JCECE Engineering JCECE Engineering Solved Paper-2011

  • question_answer
    Two sources \[A\] and \[B\] are sounding notes of frequency\[680\,\,Hz\]. A listener moves from \[A\] to \[B\] with a constant velocity\[u\]. If the speed of sound\[340\,\,m/s\], what must be the value of \[u\] so that he hears \[10\] beats per second?

    A) \[2.0\,\,m/s\]                   

    B) \[2.5\,\,m/s\]

    C)  \[3.0\,\,m/s\]                  

    D)  \[3.5\,\,m/s\]

    Correct Answer: B

    Solution :

    Apparent frequency due to source\[A\]                 \[n'=\frac{v-{{v}_{0}}}{v}\]                     \[=\frac{v-4}{v}\times n\] Apparent frequency due to source\[B\]                 \[n''=\frac{v+u}{v}\]                 \[=\frac{v-u}{v}\times n\] \[\therefore \]  \[n''-n'=\frac{2u}{v}\times n={{I}_{0}}\]                 \[u=\frac{10u}{2n}\]                 \[=\frac{10\times 340}{2\times 680}\]                 \[=2.5\,\,m/s\]


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