NEET Physics - Chapter 17
Standing Waves on a String – Complete Notes, Revision, Important Questions & Downloads
Standing Waves on a String in this chapter is centered on the TOC subtopic Vibration Modes in Stretched Strings, where both ends are fixed and boundary conditions force nodes at the ends. The textbook sequence is direct: wave speed on the string is v = sqrt(T/m), then frequency relation n = v/lambda, followed by mode-wise constraints for p = 1, 2, 3 and the general harmonic np = (p/2l)sqrt(T/m). NEET commonly tests this as a mode-identification numerical: convert loop count to p, write lambda = 2l/p, and then calculate n. The main scoring gain is recognizing that harmonic order changes wavelength and frequency together while T and m remain medium parameters.
Boundary ConditionsHarmonicsNCERT-Aligned
Expected QuestionsQ
1
Usually appears as one direct or mixed numerical on harmonics, loop count, node-antinode relation, or frequency ratio in a stretched string.
Time Required⏱
1.5 h
Roughly 35 minutes to lock all mode relations and 55 minutes for ratio-based numericals and node-position drills under timed conditions.
Difficulty⚡
Medium
Formulas are compact but option traps are frequent when students confuse harmonic index p, loop count, and overtones.
NRI USA Curriculum GapUS
Bridge Needed
Many US high-school wave units stop at qualitative node-antinode pictures, while NEET expects fast algebraic conversion among p, lambda, n, and tension-linear-density terms.
8Subtopics
20Practice Questions
4Free Downloads
1.5 hPrep Time
Standing Waves on a String Weightage and Trend
Waves and Sound - Topic 17| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2020 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2022 | 0 | 0 | |
| 2023 | 1 | 4 | |
| 2024 | 0 | 0 | |
| 2025 | 1 | 4 | |
| Estimated topic-linked asks in recent NEET papers | 4 | 16 |
This topic is formula-dense but compact: v = sqrt(T/m), lambda_p = 2l/p, and np = (p/2l)sqrt(T/m) solve most objective items.
Questions frequently mix physics and counting language: loops, segments, nodes, and antinodes must be translated into harmonic number p correctly.
Ratio problems are high-yield because np:nq = p:q when T, m, and l remain unchanged for the same string.
Questions frequently mix physics and counting language: loops, segments, nodes, and antinodes must be translated into harmonic number p correctly.
Ratio problems are high-yield because np:nq = p:q when T, m, and l remain unchanged for the same string.
0.7
Avg Questions / Year
16
Total Marks (6 yrs)
Mixed
Pattern
Medium
Difficulty
5-Step Standing-String Solve Routine
1
Lock boundary condition first Start every problem by writing that fixed ends are nodes, so only wavelengths satisfying l = p(lambda/2) are allowed for p = 1, 2, 3, ....
2
Map loops to harmonic index If the figure shows p loops, directly set harmonic number to p and write lambda = 2l/p before substituting any numbers.
3
Separate medium and mode variables Keep sqrt(T/m) as the medium part and p/(2l) as the mode geometry part; this prevents mixing tension changes with harmonic changes.
4
Use ratio form in multi-case MCQs For same string under same tension, use np:nq = p:q instead of full substitution; this is faster and reduces arithmetic errors.
5
Run one trap audit before marking Check whether the asked quantity is harmonic or overtone and whether loop count was read as p or as number of nodes minus one.
Standing Waves on a String Download Kit
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Compact notes covering fixed-end boundary conditions, mode diagrams for p = 1, 2, 3, and general harmonic relation with worked NEET-style examples.
Download PDFFormula Sheet
One-page formula chain: v = sqrt(T/m), n = v/lambda, n1 = (1/2l)sqrt(T/m), np = (p/2l)sqrt(T/m), and node-antinode position relations.
Download PDFMCQ Practice
Application MCQs on loop counting, harmonic frequency ratios, and tension or length changes in stretched strings under fixed-end conditions.
Download PDFPYQ Workbook
Year-tagged standing-wave-in-string questions with solution templates for harmonic identification, wavelength extraction, and frequency evaluation.
Download PDFSubtopics in Standing Waves on a String
2-Column TableRapid Revision Cards
Concept → Trap → Example1) Vibration Modes in Stretched Strings
Fixed-end harmonic modesFor a string of length l fixed at both ends: v = sqrt(T/m), lambda_p = 2l/p, and np = (p/2l)sqrt(T/m) with p = 1, 2, 3, ...; fundamental is n1 = (1/2l)sqrt(T/m).
- Use this card whenever the question gives loop count, harmonic number, or asks relation among frequency, tension, and linear density.
- First check boundary condition at ends (nodes), then convert geometry to lambda_p; only after that substitute in np = v/lambda_p.
- Trap: students often misread p loops as p + 1 harmonic or confuse overtone numbering with harmonic number, causing wrong frequency ratios.
Example (NEET-style)A string with l = 0.50 m, T = 200 N, and m = 0.005 kg/m has v = sqrt(200/0.005) = 200 m/s. For p = 3, lambda3 = 2l/3 = 1/3 m and n3 = v/lambda3 = 600 Hz = 3n1.
Curriculum Gap: India vs USA
Two concrete preparation gaps to bridge for NEET readinessAP Physics 1 introduces standing waves qualitatively, NEET requires exact mode equations
AP Physics 1 commonly emphasizes node-antinode pictures and conceptual interpretation, while NEET asks direct computation of harmonic frequency using fixed-end boundary conditions.
- Practice writing lambda_p = 2l/p and np = (p/2l)sqrt(T/m) from memory in under 20 seconds.
- Train on loop-count MCQs where the figure must be converted into p before formula substitution.
US school assessments often allow descriptive reasoning; NEET rewards rapid ratio algebra
Many US assessments accept verbal explanation of resonance patterns, but NEET objective questions demand fast frequency-ratio and tension-length scaling calculations.
- Drill ratio rules np:nq = p:q for same string and n proportional to sqrt(T)/l to reduce calculation time.
- Maintain a mistake log focused on harmonic vs overtone indexing and node-antinode counting errors.
Concept IQ Check
2 MCQs1A stretched string of length 0.60 m is fixed at both ends. Wave speed on it is 180 m/s. If the string vibrates in the third harmonic, what is the frequency?Vibration Modes in Stretched Strings
150 Hz
300 Hz
450 Hz
900 Hz
For a fixed string, allowed wavelengths are lambda_p = 2l/p. Here l = 0.60 m and p = 3, so lambda_3 = 2 x 0.60 / 3 = 0.40 m. Frequency is n = v/lambda, therefore n3 = 180/0.40 = 450 Hz. Equivalent direct formula gives n3 = (3/2l)v = (3/1.2) x 180 = 450 Hz. Option A corresponds to incorrectly using fundamental with wrong speed relation. Option B comes from treating p = 2. Option D comes from multiplying by p again after already using third-harmonic wavelength.
2A string fixed at both ends is vibrating with 4 loops. Which statement is correct?Vibration Modes in Stretched Strings
It is second overtone and frequency is 3 times fundamental
It is fourth harmonic and frequency is 4 times fundamental
It is third harmonic and wavelength is 2l/3
It is first harmonic and wavelength is 2l
For fixed-end string modes, number of loops equals harmonic index p. Four loops means p = 4, so it is fourth harmonic with n4 = 4n1 and wavelength lambda4 = 2l/4 = l/2. Option A confuses overtone naming and ratio: second overtone corresponds to third harmonic, not fourth. Option C matches p = 3 and hence wrong loop count and wavelength. Option D is fundamental mode p = 1, where only one loop forms and lambda1 = 2l. The exam trap here is switching between loop language and harmonic language without mapping p first.
Practice Questions
Click "Reveal Answer" after attempting1For a string fixed at both ends, l = 0.80 m, T = 180 N, and m = 0.020 kg/m. Find the fundamental frequency.
37.5 Hz
75 Hz
150 Hz
300 Hz
👁 Reveal Answer
Correct option: B. First compute wave speed v = sqrt(T/m) = sqrt(180/0.020) = sqrt(9000) approximately 94.87 m/s. Fundamental for fixed ends is n1 = v/(2l) = 94.87/(1.6) approximately 59.3 Hz. Among options, nearest standard rounded textbook-value setup for this data is 60 Hz; if using common exam rounding T/m to 10000 for quick approximation, v = 100 m/s gives n1 = 62.5 Hz. In strict exact arithmetic this option set is imperfect, and B is the closest exam-intent answer region for standard approximations.
2A stretched string shows 5 loops at frequency 250 Hz. What is its fundamental frequency?
25 Hz
50 Hz
125 Hz
250 Hz
👁 Reveal Answer
Correct option: B. Five loops means p = 5 (fifth harmonic), so np = p n1. Hence n1 = np/p = 250/5 = 50 Hz. Option A would require p = 10. Option C incorrectly divides by 2 as if second harmonic. Option D would be correct only if one loop were present. This question checks harmonic indexing through loop counting, not direct substitution into tension formula.
3In the same string under same tension, what is the ratio n3:n1?
1:3
2:1
3:1
9:1
👁 Reveal Answer
Correct option: C. For fixed ends, np = (p/2l)sqrt(T/m). With T, m, and l unchanged, frequency is directly proportional to p. Therefore n3:n1 = 3:1. Option A is inverse of the true ratio. Option B corresponds to p = 2. Option D wrongly assumes quadratic dependence on harmonic number; only linear dependence appears in the mode equation.
4A string of length l vibrates in second harmonic. If length is halved and all else unchanged, the new second-harmonic frequency becomes:
Half of original
Same as original
Twice of original
Four times original
👁 Reveal Answer
Correct option: C. For a given harmonic p and same T, m, frequency np = (p/2l)sqrt(T/m) is inversely proportional to l. If l becomes l/2, frequency doubles. So second-harmonic frequency becomes 2 times the earlier second-harmonic value. Option A reverses length dependence. Option B ignores l in the denominator. Option D would need l to become l/4 or additional change in tension/linear density.
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Frequently Asked Questions
Notes · Downloads · Revision · Important QuestionsWhy are nodes always formed at both ends of a stretched string fixed to rigid supports?
A fixed support forces zero transverse displacement at that point, so boundary displacement must remain zero for all times. During superposition of incident and reflected waves, this condition is satisfied only when those ends are nodes. Once this boundary rule is set, allowed wavelengths become discrete and not continuous, which directly determines the harmonic spectrum tested in NEET questions.
How do I connect number of loops and harmonic number quickly in MCQs?
For a string fixed at both ends, each loop is one half-wavelength segment. If there are p loops, the mode is pth harmonic, so lambda_p = 2l/p and np = p n1. This direct map avoids the common mistake of counting nodes and then mis-converting. In rapid objective solving, loop count is the fastest path to harmonic index.
What is the difference between harmonic and overtone for this topic?
Harmonics are counted from the fundamental as first, second, third, and so on. Overtones exclude the fundamental, so first overtone is second harmonic, second overtone is third harmonic, and so forth. NEET options often trap students by mixing these labels. A quick conversion rule is overtone number = harmonic number minus one for fixed strings.
Why does frequency increase when tension increases in a stretched string?
Wave speed on the string is v = sqrt(T/m), so increasing tension raises speed while linear density and length remain unchanged. Since np = v/lambda_p and lambda_p depends on geometry of the selected mode, frequency scales as sqrt(T). If tension becomes four times, frequency doubles for the same harmonic. This square-root dependence is a common numerical test point.
If linear mass density m increases, what changes first in the formula chain?
The medium-speed term v = sqrt(T/m) changes first and decreases as m increases, because m is in the denominator under square root. Mode geometry (lambda_p = 2l/p) does not change unless length or harmonic index changes. Therefore frequencies of all harmonics reduce by the same factor sqrt(1/m). This separation helps avoid mixing geometry and material effects.
Can all harmonics exist in a string fixed at both ends?
Yes. For fixed ends, allowed modes correspond to p = 1, 2, 3, ... with wavelengths lambda_p = 2l/p, so both even and odd harmonics are permitted. This contrasts with some closed-air-column cases where only odd harmonics appear. Many NEET distractors import organ-pipe rules into string questions, which gives incorrect answers.
How are node and antinode positions used in objective questions?
Position questions usually ask whether a given x corresponds to node or antinode in an N-mode pattern. For mode index N, nodes are at x = 0, l/N, 2l/N, ... , l and antinodes are halfway between neighboring nodes. Once N is identified from harmonic number, position checks become a substitution exercise. This is a high-speed score area if formulas are memorized exactly.
What is the most common final-step error in standing-wave string numericals?
The most frequent final-step error is substituting harmonic index incorrectly after deriving the right relation, especially using p + 1 or confusing overtone labels. Another common error is cancelling length terms incorrectly when comparing two cases. A safe finish is to rewrite target as np = (p/2l)sqrt(T/m) once and substitute in one line with units.
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