Answer:
Three other length for which resonance will occur is 66 cm, 110 cm and 154 cm
Explanation:
As we know that the length of the pipe is 22 cm
Now let say it is in 1st harmonic state so we have
[tex]\frac{\lambda}{4} = L[/tex]
so we have
[tex]\frac{\lambda}{2} = 2L[/tex]
So we have
[tex]L_1 = L + \frac{\lambda}{2}[/tex]
[tex]L_1 = L + 2L[/tex]
[tex]L_1 = 22 + 2L[/tex]
For N = 1
[tex]L_1 = 66 cm[/tex]
For N = 2
[tex]L_2 = 22 + 88[/tex]
[tex]L_2 = 110 cm[/tex]
For N = 3
[tex]L_3 = 22 + 132[/tex]
[tex]L_3 = 154 cm[/tex]