Two wires are fixed in a sonometer. Their tensions are in the ratio $8 : 1$ . The lengths are in the ratio $36 : 35$ . The diameters are in the ratio $4 : 1$ . Densities of the materials are in the ratio $1 : 2$ . If the higher frequency in the setting is $360 H z$ , the beat frequency when the two wires are sounded together, is:

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Solution

The correct option is C 10
$n = \frac{1}{2 l} \sqrt{\frac{T}{m}}$

$= \frac{1}{2 l} \sqrt{\frac{T}{π r^{2} ρ}}$

$= \frac{1}{2 l r} \sqrt{\frac{T}{π ρ}}$

$\frac{n_{1}}{n_{2}} = \frac{l_{2}}{l_{1}} . \frac{r_{2}}{r_{1}} . \sqrt{\frac{T_{1}}{T_{2}}} . \sqrt{\frac{ρ_{2}}{ρ_{1}}}$

$= \frac{35}{36} . \frac{1}{4} . \sqrt{\frac{8}{1}} . \sqrt{\frac{2}{1}}$

$= \frac{35}{36}$

Given: Higher frequency, $n_{2} = 360 H z$

$\Rightarrow n_{1} = \frac{35}{36} \times 360 = 350$

Therefore, beat frequency $= 360 - 350 = 10 H z$

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Two wires are fixed in a sonometer. Their tensions are in the ratio 8 : 1. The lengths are in the ratio 36 : 35. The diameters are in the ratio 4 : 1. Densities of the materials are in the ratio 1 : 2. If the lower frequency in the setting is 360 Hz. the beat frequency when the two wires are sounded together is

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The correct option is C 10n=12l√Tm =12l√Tπr2ρ =12lr√Tπρn1n2=l2l1.r2r1.√T1T2.√ρ2ρ1 =3536.14.√81.√21 =3536Given: Higher frequency, n2=360 Hz⇒n1=3536×360=350Therefore, beat frequency =360−350=10 Hz