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Consider a strong connected At both ends with rigid wall (Fig. 16-36). Find the second longest resonance wave length for which 3/8 is an antinode. |
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Answer» Solution :If we use normal frequencies for string with both rigid ends we can solve this problem simply by trial and error. We can consider rope of length 31/8 with one end rigid and one end free. Also, we can consider a rope of length I with both ends rigid. The frequencies which SATISFY both equations SIMULTANEOUSLY are the required frequencies. Calculations. All possible frequencies which will satisfy condition of both ends rigid: `f_(1)=m/(2F) sqrt(T/mu)` All possible frequencies which will satisfy condition of rope of length 3/8 with one end rigid and one end free : `f_(2)=(2n+1)/(4(3!//8)) sqrt(T/mu)` Only those vibrations will have resonance which satisfy `f_(1)=f_(2)`, so `(2n+1)/(4(3//8)) sqrt(T/mu) =f= (m)/(2f) sqrt((T)/(mu))` Therefore `8n+4=3m` 3m-8n=4 `m""n` `4"1"` `"124"` `"207"` The second row has the second lowest frequency, hence second largest WAVELENGTH. Therefore, on substituting value, `lambda=6t`. We can verify the answer by checking for the above value that results, in satisfying both the conditions. |
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