If the first position of resonance in a resonance tube is 18.0cm from the open end, the distance from the open end to the next position of resonance will be

Analyze the question: In a resonance tube (a tube closed at one end, with the water level adjusted), the first resonance happens at length L₁ = λ/4 (ignoring end correction). We are asked for the distance of the next resonance from the open end.

Concept:

• First resonance: L₁ = λ/4
• Second resonance: L₂ = 3λ/4
• Third resonance: L₃ = 5λ/4

The difference between successive resonance positions is λ/2.

Solution:

Take L₁ = 18.0 cm = λ/4 (neglecting end correction).

Then λ = 72 cm.

L₂ = 3λ/4 = 3 × 18.0 = 54.0 cm

Equivalently, L₂ = L₁ + λ/2 = 18 + 36 = 54 cm.

Answer: The next position of resonance is 54.0 cm from the open end, which is option E.

Note on end correction: If an end correction e were included, L₁ + e = λ/4 and L₂ + e = 3λ/4, so L₂ = 3L₁ + 2e. Since no end correction is given, we take e = 0, so L₂ = 54 cm.

Exam tip: In a resonance tube, the resonance lengths are in the ratio 1 : 3 : 5. If the first is 18 cm, the next is 3 × 18 = 54 cm and the one after is 5 × 18 = 90 cm. Remember "odd multiples of λ/4."