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BVP Medical BVP Medical Solved Paper-2006

  • question_answer
    Standing waves are produced in a 10 m long stretched string. If the string vibrates in 5 segments and the wave velocity is 20 m/s, the frequency is :

    A) 10 Hz                                    

    B)  5 Hz

    C)  4 Hz                                      

    D)  2 Hz

    Correct Answer: B

    Solution :

                    In the case of standing wave, the length of one segment is \[\frac{\lambda }{2}\]. There are 5 segments and total length of string is 10 m. \[\therefore \]  \[5\frac{\lambda }{2}=10\] \[\Rightarrow \]               \[\lambda =4m\] Frequency, \[n=\frac{v}{\lambda }=\frac{20m/s}{4m}=5Hz\] NOTE: Standing wave is an example of interference. Destructive interference means node and constructive interference means antinode.


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