Given below are two statements: Statement I: In a stationary wave, the distance between two adjacent nodes is one full wavelength. Statement II: An air column closed at one end supports only the odd harmonics of its fundamental. Choose the correct answer.

Statement I is incorrect but Statement II is correct

Statement I is correct but Statement II is incorrect

Both Statement I and Statement II are correct

Both Statement I and Statement II are incorrect

Waves

6 NEET-UG practice questions on Waves, part of Physics. The first 6 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Easy
Given below are two statements: Statement I: In a stationary wave, the distance between two adjacent nodes is one full wavelength. Statement II: An air column closed at one end supports only the odd harmonics of its fundamental. Choose the correct answer.
1. AStatement I is correct but Statement II is incorrect
2. BBoth Statement I and Statement II are correct
3. CStatement I is incorrect but Statement II is correct
4. DBoth Statement I and Statement II are incorrect
Show answerHide answer
Answer: C · Statement I is incorrect but Statement II is correct
Adjacent nodes are λ/2 apart, so Statement I is false. A pipe closed at one end has a node at that end and an antinode at the open end, so only f, 3f, 5f and so on can fit, and Statement II is true.
- A: This reverses the truth values of the two statements.
- B: Statement I is false, since the spacing is λ/2.
- D: Statement II is true: the closed pipe has only odd harmonics.
Remember this
Node to node is λ/2; node to adjacent antinode is λ/4.
Question 2 · Medium
A string 1 m long of linear mass density 0.01 kg/m is fixed at both ends and kept under a tension of 100 N. What is the frequency of its third harmonic?
1. A300 Hz
2. B150 Hz
3. C100 Hz
4. D50 Hz
Show answerHide answer
Answer: B · 150 Hz
Wave speed v = √(T/μ) = √(100/0.01) = 100 m/s. The fundamental is v/(2L) = 50 Hz, so the third harmonic is 3 × 50 = 150 Hz.
- A: This uses v/L as the fundamental instead of v/(2L).
- C: This is the second harmonic.
- D: This is the fundamental, not the third harmonic.
Remember this
For a string fixed at both ends, fₙ = nv/(2L), with all harmonics present.
Question 3 · Hard
In a resonance tube experiment with a 500 Hz tuning fork, the first resonance occurs at an air-column length of 16 cm and the second at 50 cm. What are the speed of sound and the end correction?
1. A340 m/s and 2 cm
2. B333 m/s and 1 cm
3. C320 m/s and 0 cm
4. D340 m/s and 1 cm
Show answerHide answer
Answer: D · 340 m/s and 1 cm
The two lengths differ by λ/2: 50 − 16 = 34 cm, so λ = 68 cm and v = 500 × 0.68 = 340 m/s. The end correction e = (l₂ − 3l₁)/2 = (50 − 48)/2 = 1 cm. Check: l₁ + e = 17 cm = λ/4.
- A: This omits the factor 1/2 in the end correction.
- B: This uses v = (4/3)fl₂, ignoring the end correction.
- C: This uses v = 4fl₁, ignoring the end correction.
Remember this
Using the difference of two resonance lengths removes the end correction from the speed.
Question 4 · Easy
A wave on a string is given by y = 0.02 sin(4πt − 0.5πx), in SI units. What is its speed?
1. A4 m/s
2. B8 m/s
3. C2 m/s
4. D0.125 m/s
Show answerHide answer
Answer: B · 8 m/s
Speed = ω/k = 4π/(0.5π) = 8 m/s. Check: f = 2 Hz and λ = 2π/k = 4 m, so fλ = 8 m/s.
- A: This is the wavelength, 2π/k.
- C: This is the frequency, ω/2π.
- D: This inverts the ratio, using k/ω.
Remember this
For y = a sin(ωt − kx), v = ω/k, with the wave moving towards +x.
Question 5 · Medium
Match List-I with List-II. The figure draws displacement patterns in four air columns, each 85 cm long, with the speed of sound 340 m/s. List-I (Pattern) A. Pipe A B. Pipe B C. Pipe C D. Pipe D List-II (Frequency) I. 100 Hz II. 200 Hz III. 300 Hz IV. 400 Hz
Figure: MedExams
1. AA-III, B-II, C-I, D-IV
2. BA-I, B-II, C-III, D-IV
3. CA-III, B-I, C-II, D-IV
4. DA-IV, B-II, C-I, D-III
Show answerHide answer
Answer: A · A-III, B-II, C-I, D-IV
For a closed pipe f₁ = v/4L = 340/3.4 = 100 Hz, and only odd harmonics occur: C is 100 Hz and A, with L = 3λ/4, is 300 Hz. For an open pipe f₁ = v/2L = 200 Hz: B is 200 Hz and D, the second harmonic, is 400 Hz.
- B: Exchanges the closed-pipe third harmonic and fundamental.
- C: Exchanges the fundamentals of the open and closed pipes.
- D: Exchanges the patterns of pipes A and D.
Remember this
Closed pipe: L = (2n − 1)λ/4 with odd harmonics. Open pipe: L = nλ/2 with all harmonics.
Question 6 · Medium
A snapshot of a transverse wave of frequency 50 Hz on a stretched string is plotted. If the string has linear mass density 0.01 kg/m, find its tension.
Figure: MedExams
1. A1 N
2. B4 N
3. C0.2 N
4. D16 N
Answer and explanation in LearnFree account

Waves

Given below are two statements: Statement I: In a stationary wave, the distance between two adjacent nodes is one full wavelength. Statement II: An air column closed at one end supports only the odd harmonics of its fundamental. Choose the correct answer.

A string 1 m long of linear mass density 0.01 kg/m is fixed at both ends and kept under a tension of 100 N. What is the frequency of its third harmonic?

In a resonance tube experiment with a 500 Hz tuning fork, the first resonance occurs at an air-column length of 16 cm and the second at 50 cm. What are the speed of sound and the end correction?

A wave on a string is given by y = 0.02 sin(4πt − 0.5πx), in SI units. What is its speed?

Match List-I with List-II. The figure draws displacement patterns in four air columns, each 85 cm long, with the speed of sound 340 m/s. List-I (Pattern) A. Pipe A B. Pipe B C. Pipe C D. Pipe D List-II (Frequency) I. 100 Hz II. 200 Hz III. 300 Hz IV. 400 Hz

A snapshot of a transverse wave of frequency 50 Hz on a stretched string is plotted. If the string has linear mass density 0.01 kg/m, find its tension.