ii) Two sources of sound are producing waves of frequency n, and n2, where
(12-n2) is small, show mathematically that the beat frequency is (ny -n2).
Answers
Explanation:
Solution :-
**refer to attached for the reference.
First of all we will add some more information in the diagram in order to solve it.
As In AMG , A battery is connected we will take a current i_1i
1
Then according to Kirchoff's junction law at junction G.
i_1 = I + xi
1
=I+x
x = i_1 - Ix=i
1
−I
So current in GB = i_1 - Ii
1
−I
Now in FED we will take another current i_2i
2
.
I = i_2 + yI=i
2
+y
y = I - i_2y=I−i
2
So current in FC = I - i_2I−i
2
Now by applying junction law at other junctions
▪️Current in CB = II
▪️Current in BA = i_1i
1
Now we will write Loop For (anticlockwise)
(1)AHGB
-12 = 3R(i_1 - I) + R(i_1)−12=3R(i
1
−I)+R(i
1
)
\implies -12 = 3Ri_1 - 3RI + Ri_1⟹−12=3Ri
1
−3RI+Ri
1
\implies -12 = 4Ri_1 - 3RI⟹−12=4Ri
1
−3RI ...(i)
(2) AHFC
-12 = 4R(I - i_2) + R(i_1)−12=4R(I−i
2
)+R(i
1
)
\implies -12 = 4RI - 4Ri_2 + Ri_1⟹−12=4RI−4Ri
2
+Ri
1
...(ii)
(3) AHED
24 - 12 = 2R(i_2) + R(i_1)24−12=2R(i
2
)+R(i
1
)
\implies12 = 2Ri_2 + Ri_1⟹12=2Ri
2
+Ri
1
....(iii)
(4)CFED
24 = 2R(i_2) - 4R(I-i_2)24=2R(i
2
)−4R(I−i
2
)
\implies 24 = 2Ri_2 - 4RI + 4Ri_2⟹24=2Ri
2
−4RI+4Ri
2
\implies 24 = 6Ri_2 - 4RI⟹24=6Ri
2
−4RI ....(iv)
Now 3(iv) - 4(i)
72 = 18Ri_2 - 12RI72=18Ri
2
−12RI
+ 48 = - 16Ri_1 + 12RI+48=−16Ri
1
+12RI
________________________
120 = 18Ri_2 - 16Ri_1120=18Ri
2
−16Ri
1
.....(v)
________________________
Now from (v) - 9(iii)
120 = 18Ri_2 - 16Ri_1120=18Ri
2
−16Ri
1
-108= - 18Ri_2 - 9Ri_1−108=−18Ri
2
−9Ri
1
________________________
12 = -25Ri_112=−25Ri
1
________________________
Ri_1 = \dfrac{-12}{25}Ri
1
=
25
−12
Now from by substiting value of Ri_1Ri
1
in (i)
-12 = 4Ri_1 - 3RI−12=4Ri
1
−3RI
\implies -12 = 4\dfrac{-12}{25} - 3RI⟹−12=4
25
−12
−3RI
\implies -12 = \dfrac{-48}{25}- 3RI⟹−12=
25
−48
−3RI
\implies 3RI = \dfrac{-48}{25} + 12⟹3RI=
25
−48
+12
\implies 3RI = \dfrac{300-48}{25}⟹3RI=
25
300−48
\implies 3RI = \dfrac{252}{25}⟹3RI=
25
252
\implies 3RI = 10.08⟹3RI=10.08
\implies RI = 3.36⟹RI=3.36
\implies I = \dfrac{3.36}{R}⟹I=
R
3.36
\implies I = \dfrac{3.36}{1000}⟹I=
1000
3.36
\implies I = 3.36 \: mA⟹I=3.36mA
So current I = 3.36 mA
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