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Home » A sonometer wire of length ‘ ‘ is in unison with a tuning fork of frequency’n’. When the vibrating length of the wire is reduced to ‘ ‘, it produces ‘ ‘ beats per second with the fork. The frequency of the fork isA. B. C. D.
A sonometer wire of length ‘ \mathrm{L}_{1} ‘ is in unison with a tuning fork of frequency’n’. When the vibrating length of the wire is reduced to ‘ \mathrm{L}_{2} ‘, it produces ‘ \mathrm{x} ‘ beats per second with the fork. The frequency of the fork is
A. \frac{L_{2} x}{L_{2}-L_{1}}
B. \frac{L_{1} x}{L_{1}-L_{2}}
C. \frac{L_{2} x}{L_{2}-L_{1}}
D. \frac{L_{2} x}{L_{1}-L_{2}}
Physics
A sonometer wire of length ‘ \mathrm{L}_{1} ‘ is in unison with a tuning fork of frequency’n’. When the vibrating length of the wire is reduced to ‘ \mathrm{L}_{2} ‘, it produces ‘ \mathrm{x} ‘ beats per second with the fork. The frequency of the fork is
A. \frac{L_{2} x}{L_{2}-L_{1}}
B. \frac{L_{1} x}{L_{1}-L_{2}}
C. \frac{L_{2} x}{L_{2}-L_{1}}
D. \frac{L_{2} x}{L_{1}-L_{2}}

Correct answer is D.

For the length L_{1}, n_{1}=\frac{1}{2 L_{1}} \sqrt{\frac{T}{m}}=n \ldots(1)

where m= mass per unit length

For the length L_{2}, n_{2}=\frac{1}{2 L_{2}} \sqrt{\frac{T}{m}} \ldots . .(2)

\therefore \frac{N_{2}}{n_{1}}=\frac{L_{1}}{L_{2}} \ldots .

but L_{2}<L_{1} \therefore N_{2}>n_{1}

\therefore n_{2}-n_{1}=x or n_{2}=n_{1}+x \ldots+ \therefore \quad form (3), \frac{n_{1}+x}{n_{1}}=\frac{L_{1}}{L_{2}}

\therefore 1+\frac{x}{n_{1}}=\frac{L_{1}}{L_{2}} \therefore \frac{x}{n_{1}}=\frac{L_{1}}{L_{2}}-1=\frac{L_{1}-L_{2}}{L_{2}} \therefore n_{1}=\frac{x L_{2}}{L_{1}-L_{2}}

\therefore The frequency of the fork n=n_{1}=\left(\frac{L_{2}}{L_{1}-L_{2}}\right) x

i

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