1.

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.

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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