Question

. The Length of air column for fundamental mode in resonance tube is 16 cm and that
for the second resonance further is 50.25 cm. Find the end correction and inner
diameter of tube.​

Answer 1

Explanation:

Given  The Length of air column for fundamental mode in resonance tube is 16 cm and that  for the second resonance further is 50.25 cm. Find the end correction and inner  diameter of tube.  In resonance tube, anti node is formed a little above the open end called as end correction.

• So l1 + e = λ / 4 and l2 + e = 3λ / 4
• So e = l2 – 3 l1 / 2
• So l1 = 16 cm and l2 = 50.25 cm
• Now e = 50.25 – 3(16) / 2
•           = 2.25 / 2
• So e = 1.125 cm
• Now e = 0.3 d
• So d = e / 0.3
• Or d = 1.125 / 0.3

Or d = 3.75 cm

Answer 2

Thus the diameter of the tube is d = 3.75 cm

Explanation:

We are given that:

• Length of column "l1" = 16 cm
• Length of second resonance "l2" = 50.25 cm
• To Find: End correction "e" and diameter "d" = ?

Solution:

e = l - 3 l1 / 2

e = 50.25 - (3 x 16) / 2

e = 50.25 - 48 / 2

e = 2.25 / 2 = 1.125 cm

e = 0.3 d

Now diameter:

d = e / 0.3 = 1.125 / 0.3 = 3.75 cm

Thus the diameter of the tube is d = 3.75 cm