Harmonics on strings and in pipes: Class 11 practice
A string fixed at both ends and an open pipe have fₙ = nv/2L (all harmonics). Read the method, follow the three worked examples, then try the six questions.
What you should know first
- v = fλ
The idea behind harmonics on strings and in pipes
A string fixed at both ends and an open pipe have fₙ = nv/2L (all harmonics). A pipe closed at one end has fₙ = nv/4L with only odd n.
Step-by-step method
- Identify the system (string/open pipe/closed pipe).
- Write the fundamental: v/2L or v/4L.
- Multiply by n (odd n only for closed pipes).
Worked example 1 (Basic)
Find the fundamental frequency of an open pipe 0.99 m long (v = 330 m/s).
- f₁ = v/2L = 330/1.98
- = 166.67 Hz
- Answer: 166.67 Hz
Worked example 2 (Standard)
Find the first two frequencies of a pipe 0.4 m long closed at one end (v = 340 m/s).
- f₁ = v/4L = 212.5 Hz
- Next allowed: 3f₁ = 637.5 Hz
- Answer: 212.5 Hz and 637.5 Hz
Worked example 3 (Challenge)
A 0.61 m string with tension 125 N and mass per length 0.007 kg/m vibrates in its fundamental. Find the frequency.
- v = √(T/μ) = 133.63 m/s
- f = v/2L = 109.53 Hz
- Answer: 109.53 Hz
Common mistakes and misconceptions
- Using even harmonics in a closed pipe
- Using 2L for a closed pipe
Practice questions
- (Basic) Find the fundamental frequency of an open pipe 0.88 m long (v = 330 m/s).
- (Basic) Find the fundamental frequency of an open pipe 0.61 m long (v = 330 m/s).
- (Standard) Find the first two frequencies of a pipe 0.37 m long closed at one end (v = 340 m/s).
- (Standard) Find the first two frequencies of a pipe 0.46 m long closed at one end (v = 340 m/s).
- (Challenge) A 0.27 m string with tension 138 N and mass per length 0.005 kg/m vibrates in its fundamental. Find the frequency.
- (Challenge) A 0.43 m string with tension 154 N and mass per length 0.005 kg/m vibrates in its fundamental. Find the frequency.
Answer key with full working
- Q1: f₁ = v/2L = 330/1.76 → = 187.5 Hz → Answer: 187.5 Hz
- Q2: f₁ = v/2L = 330/1.22 → = 270.49 Hz → Answer: 270.49 Hz
- Q3: f₁ = v/4L = 229.73 Hz → Next allowed: 3f₁ = 689.19 Hz → Answer: 229.73 Hz and 689.19 Hz
- Q4: f₁ = v/4L = 184.78 Hz → Next allowed: 3f₁ = 554.35 Hz → Answer: 184.78 Hz and 554.35 Hz
- Q5: v = √(T/μ) = 166.13 m/s → f = v/2L = 307.65 Hz → Answer: 307.65 Hz
- Q6: v = √(T/μ) = 175.5 m/s → f = v/2L = 204.07 Hz → Answer: 204.07 Hz
Syllabus note
This skill is listed under "Waves" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is harmonics on strings and in pipes in?
"Waves", Class 11 Physics (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.