2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Wave Mechanics

JEE Main & Advanced Physics Wave Mechanics Question Bank Beats

  • question_answer
    A sound source of frequency 170 Hz is placed near a wall. A man walking from a source towards the wall finds that there is a periodic rise and fall of sound intensity. If the speed of sound in air is 340 m/s the distance (in metres) separating the two adjacent positions of minimum intensity is [MNR 1992; UPSEAT 2000; CPMT 2002]

    A)            1/2                                           

    B)            1

    C)            3/2

    D)            2

    Correct Answer: B

    Solution :

               \[v=n\lambda \]Þ \[\lambda =\frac{v}{n}=\frac{340}{170}\]Þ \[\lambda =2\] Distance separating the position of minimum intensity = \[\frac{\lambda }{2}=\frac{2}{2}=1\]m


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