2. Solved Papers
  3. JIPMER

JIPMER Jipmer Medical Solved Paper-2003

  • question_answer
    A steel wire 0.5 m long has a total mass kg and stretched with a tension of 800 N. The frequency with which it vibrates in its fundamental note will be :

    A)  4 Hz                      

    B)         2 Hz                      

    C)  200 Hz                 

    D)         160 Hz

    Correct Answer: C

    Solution :

    Mass per unit length \[m=\frac{M}{L}=\frac{0.01}{0.5}kg\text{/}m\] Fundamental frequency is \[n=\frac{1}{2L}\sqrt{\frac{T}{m}}=\frac{1}{2\times 0.5}\sqrt{\frac{800\times 0.5}{0.01}}\]       \[=200Hz\]


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