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A string, fixed at both ends, vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6 cm. Find the length of the string. |
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Answer» Correct Answer - C Let there be n loops in the 1st case `rarr` Length of the wire` `L =((n`lamda_1`)/2) (`lamda_1`=`2xx2=4cm`) `rarr` Length of the wire `L_l={(n+1)lamda_2/2}` `(lamda_2=2xx(1.6)=3.2cm)` `rarr (nlamda_1)/2=(n+1)lamda_2/2` `rarr nxx4=(n+1)(3.2)` `rarr 4n-(3.2)n=3.2` `rarr 0.8n=3.2` `rarr n=4` `:. ` Length of the string `L=(n lamda_1)/2=((4xx4))/2=8 cm` |
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