First slide

Stationary waves

Question

In a resonance pipe the first and second resonances are obtained at depths 22.7 cm and 70.2 cm respectively. What will be the end correction

Moderate

A

1.05 cm
B

115.5 cm
C

92.5 cm
D

113.5 cm

Solution

For end correction x, $\frac{l_{2} + x}{l_{1} + x} = \frac{3 λ / 4}{λ / 4} = 3$
$x = \frac{l_{2} - 3 l_{1}}{2} = \frac{70 .2 - 3 \times 22 .7}{2} = 1 .05 cm$

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