Aditi Chauhan
Last Activity: 10 Years ago
Hello Student,
Please find the answer to your question
It is given that C as a node. This implies that at A and B antinodes are formed. Again it is given that the frequencies are same.
v
1/4ℓ x p
1 = v
2/4ℓ x p
2 or p
1/p
2 = v
1/v
2 3/11
Or, 11p1 = 3p2
This means that the third harmonic in AC is equal to 11th harmonic in CB.
Now, the fundamental frequency in AC
= v1 4ℓ = 1100/4 x 0.5 = 550Hz
And the fundamental frequency in CB
= v2/4ℓ = 300/4 x 0.5 = 550 Hz
∴ Frequency in AC = 3 x 550 = 1650 Hz
And frequency in CB = 11 x 150 = 1650Hz.
Thanks
Aditi Chauhan
askIITians Faculty