$5$ beats/ second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either $0.95 m$ or $1 m$ . The frequency of the fork will be:

195 H z

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251 H z

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150 H z

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300 H z

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Solution

The correct option is A $195 H z$
$L_{1} = 0.95 m, L_{2} = 1 m$
$L_{2} > L_{1}, n_{1} > N > n_{2}$
$n_{1} - N = 5 a n d N - n_{2} = 5$
on solving $n_{1} - n_{2} = 10$
$n_{2} = n_{1} - 10$
By law of length of vibrating string
$n_{1} L_{1} = n_{2} L_{2}$
On solving we get $n_{1} = 200 H z$
$n_{1} - N = 5$
N = 195 Hz

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5 beats/ second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95m or 1m. The frequency of the fork will be:

The correct option is A 195HzL1=0.95m,L2=1mL2>L1,n1>N>n2n1−N=5andN−n2=5on solving n1−n2=10n2=n1−10By law of length of vibrating string n1L1=n2L2On solving we get n1=200Hzn1−N=5N = 195 Hz

$5$ beats/ second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either $0.95 m$ or $1 m$ . The frequency of the fork will be: