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PhysicsSound WaveBeatsHard2 minPYQ_2020

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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 higher frequency in the setting is 360 Hz, the beat frequency when the two wires are sounded together, is

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Answer:

Solution:

Frequency of sonometer wire is given by

$v = \frac{1}{2 l} \sqrt{\frac{T}{m}} = \frac{1}{2 l} \sqrt{\frac{T}{π r^{2} p}}$

$v_{1} = \frac{1}{2 l_{1}} \sqrt{\frac{T_{1}}{π r_{1}^{2} ρ_{1}}}$

$v_{2} = \frac{1}{2 l_{2}} \sqrt{\frac{T_{2}}{π r_{2}^{2} ρ_{2}}}$

$∴ \frac{v_{1}}{v_{2}} = \frac{l_{2}}{l_{1}} \sqrt{\frac{T_{1}}{T_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}} \times \frac{ρ_{2}}{ρ_{1}}}$

$\frac{v_{1}}{v_{2}} = \frac{35}{36} \sqrt{\frac{8}{1} \times \frac{1}{16} \times \frac{2}{1}}$

$∵ v_{1} < v_{2} a n d v_{2} = 360 H z$

Therefore,

$v = 360 \times \frac{35}{36}$

$v_{1}$ =350 Hz

So, number of beats produced = $v_{1} - v_{2}$

=360-350=10

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Beats

⏱ 2mℹ️ Source: PYQ_2020

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