if l1 and l2 are the lengths of the air column for the first and second resonance when a tuning fork of frequency n is sounded on a resonance tube, then what is the distance of the displacement antinode from the top end of the resonance tube.
Answers
The distance of the displacement antinode from the top end of the resonance tube is e [ e = l₂ - 3l₁ / 2 ]
Explanation:
=> Suppose the end correction is e.
=> For the 1st resonance:
e + l₁ = λ/4 ...(1)
=> For 2nd resonance:
e + l₂ = 3λ/4 ...(2)
=> By placing the value of λ/4 from eq (1) in eq (2), we get
e + l₂ = 3 (e + l₁)
e + l₂ = 3e + 3l₁
3e - e = l₂ - 3l₁
2e = l₂ - 3l₁
e = l₂ - 3l₁ / 2
Thus, the distance of the displacement antinode from the top end of the resonance tube is e.
Learn more:
Q:1 Explain why (or how):
(a) In a sound wave, a displacement node is a pressure antinode and vice versa,
(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,
(c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
(e) The shape of a pulse gets distorted during propagation in a dispersive medium.
Click here: https://brainly.in/question/1524699
Q:2 A tuning fork of frequency 500 Hz is sounded on a resonance tube. The first & second resonances are obtained at 17 cm and 52 cm. The velocity of sound in m/s is?
Click here: https://brainly.in/question/7381583