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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2002

  • question_answer
    Linear density of a string is \[1.3\times {{10}^{-4}}\,kg/m\] and wave equation is y = 0.021 sin (x + 30 0. Find the tension in the string where x in metre and t in second:

    A)  0.12 N                 

    B)  0.21 N

    C)  1.2 N                    

    D)         0.012 N

    Correct Answer: A

    Solution :

    Mass per unit length\[m=1.3\times {{10}^{-4}}kg/m\] \[y=0.021\sin (x+30t)\] Comparing with\[y=a\sin \left( \omega t=\frac{2\pi x}{\lambda } \right)\]                                 \[a=0.021,\omega =30\] \[\Rightarrow \]               \[n=\frac{30}{2\pi }\]                                 \[\frac{2\pi }{\lambda }=1\]                                 \[\lambda =\frac{\lambda }{2\pi }\] Frequency\[n=\frac{1}{\lambda }\frac{\sqrt{1}}{m}\]                             \[\frac{30}{2\pi }=\frac{1}{2\pi }\sqrt{\frac{T}{1.3\times {{10}^{-4}}}}\]                    \[T=1.3\times {{10}^{-4}}\times 900\]                                      \[=0.117=0.12N\]


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