Question

Ques. is

The two parts of a dominated wire divided by a movable knife edge differ by 2mm. And produce beat per second when sounded together . Find their frequencies if the whole length of the wire is 1 metre...!


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Answer 1

Hey!
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❄There's a correction in the question - instead of dominated it should be sonometer !!

Ques.→The two parts of a sonometer wire divided by a movable knife edge differ by 2mm. And produce one beat per second when sounded together . Find their frequencies if the whole length of the wire is 1 metre .

Solution →

◾Here we have to find frequencies . So let us denote the frequencies of the two parts by v1 and v2.

◾Also , let l1 and l2 denote the divided lengths of the sonometer.

➖◾Given that , the entire length is 1 m = 100 cm

•°• l1 + l2 = 100 cm_________________( 1 )

◾➖Also , their difference is given 2mm = 0.2 cm

•°• l1 - l2 = 0.2 cm___________________(2)

❄Solve Equation (1) and (2) , we will get

◼ l1 => 50.1 cm and l2 = 49.9 cm

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→ To find the frequencies v1 and v2 , we are given that they produce 1 beat per second .

•°• v2 - v1 = 1

➖◾But we need some more equation to find the two variables . So here we shall apply the law of length ,

$= > \frac{v2}{v1} = \frac{l1}{l2} = \frac{50.1}{49.9}$

So now we can substitute the values ,

$= > \frac{50.1}{49.9}v1 \: - v1 = 1 \\ \\ = > v1 = \frac{49.9}{0.2}$

{ This is just by substituting v1 from the equation and then solving }

•°• v1 = 249.5 Hz ✔

◼Put v1 in the equation (3) we have ,

=> v2 = 249.5 + 1

•°• v2 = 250.5 Hz ✔

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Answer 2

$<b>Hello Friend<b>$

Let l1 = length of first part of sonometer

l2 = length of second part of sonometer

l1 + l2 = 100cm

l1-l2= 0.2cm

on adding

we get ,,

2l1= 100.2

l1= 100.0/2

= 50.1 cm

So,,

L2 = 49.9cm

Hope helps ❤️