Physics Mechanics - How to Solve: Wave on a String (Speed, Reflection, Standing Waves, Sonometer) – IIT JEE Guide

How to Solve: Wave on a String (Speed, Reflection, Standing Waves, Sonometer) – IIT JEE Guide

Introduction

"Mastering wave speed, reflection, and standing waves on a string can fetch you 8–12 marks in IIT JEE—enough to push you into the top 1%." This topic appears in JEE Main (2–3 questions/year) and JEE Advanced (1–2 questions, often in Section B). It’s also the foundation for sound waves, interference, and musical instruments—real-world applications like guitar tuning, ultrasound imaging, and earthquake-resistant buildings.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand: 1. Basic wave terminology (amplitude, wavelength, frequency, phase). 2. Newton’s laws of motion (tension, acceleration). 3. Simple harmonic motion (SHM) (restoring force, angular frequency).

If any of these are shaky, pause and review them first.

KEY TERMS & FORMULAS

1. Wave Speed on a String

Formula: [ v = \sqrt{\frac{T}{\mu}} ] - ( v ) = wave speed (m/s) MEMORISE THIS - ( T ) = tension in the string (N) - ( \mu ) = linear mass density (kg/m) = mass/length

Why it works: Tension provides the restoring force; higher tension = faster waves. Thicker strings (higher ( \mu )) slow waves down.

2. Reflection of Waves

Key Concepts: - Fixed end: Wave reflects inverted (phase change of ( \pi ) radians). - Free end: Wave reflects upright (no phase change). - Impedance mismatch: Partial reflection + transmission when string density changes.

Formula (for partial reflection): [ R = \left( \frac{Z_2 - Z_1}{Z_2 + Z_1} \right)^2 ] - ( R ) = reflection coefficient (fraction of energy reflected) - ( Z = \sqrt{T \mu} ) = impedance of the string MEMORISE THIS

3. Standing Waves (Stationary Waves)

Key Concepts: - Formed by superposition of two identical waves traveling in opposite directions. - Nodes: Points of zero displacement (destructive interference). - Antinodes: Points of maximum displacement (constructive interference).

Formulas: 1. Fundamental frequency (1st harmonic):
[ f_1 = \frac{v}{2L} ] MEMORISE THIS
- ( L ) = length of string (m)
- ( v ) = wave speed (from ( \sqrt{T/\mu} ))

1. nth harmonic frequency:
   [ f_n = n f_1 = \frac{n v}{2L} ] MEMORISE THIS

2. ( n ) = harmonic number (1, 2, 3, ...)

3. Wavelength of nth harmonic:
   [ \lambda_n = \frac{2L}{n} ] MEMORISE THIS

4. Sonometer (Melde’s Experiment)

Key Concepts: - A sonometer is a device to study standing waves on a string under tension. - Transverse vibrations: String vibrates perpendicular to its length. - Resonance: Occurs when driving frequency matches a natural frequency of the string.

Formula (for resonance): [ f = \frac{n}{2L} \sqrt{\frac{T}{\mu}} ] MEMORISE THIS - ( f ) = resonant frequency (Hz) - ( n ) = harmonic number (1, 2, 3, ...)

STEP-BY-STEP METHOD

Step 1: Identify the Given Quantities

• Tension (( T ))? Mass density (( \mu ))? Length (( L ))? Frequency (( f ))?
• Write down all given values with units.

Step 2: Determine What’s Asked

• Wave speed? Frequency? Wavelength? Harmonic number?
• Underline the unknown in the question.

Step 3: Choose the Right Formula

• Wave speed? Use ( v = \sqrt{T/\mu} ).
• Standing waves? Use ( f_n = \frac{n v}{2L} ) or ( \lambda_n = \frac{2L}{n} ).
• Sonometer? Use ( f = \frac{n}{2L} \sqrt{T/\mu} ).

Step 4: Plug in Values and Solve

• Substitute numbers with units.
• Simplify step-by-step (show all working).

Step 5: Check for Phase Changes (Reflection)

• Fixed end? Add ( \pi ) phase shift.
• Free end? No phase shift.

Step 6: Verify Units and Reasonableness

• Speed should be in m/s, frequency in Hz.
• If ( v > 3 \times 10^8 ) m/s, you’ve made a mistake (speed of light is the limit!).

WORKED EXAMPLES

Example 1 – Basic (Wave Speed)

Question: A string of length 2 m and mass 0.04 kg is under a tension of 100 N. What is the speed of a wave on this string?

Solution: 1. Given:
- ( L = 2 ) m
- ( m = 0.04 ) kg
- ( T = 100 ) N 2. Find: ( v = ? ) 3. Formula: ( v = \sqrt{T/\mu} ) 4. Calculate ( \mu ):
[ \mu = \frac{m}{L} = \frac{0.04}{2} = 0.02 \text{ kg/m} ] 5. Plug in:
[ v = \sqrt{\frac{100}{0.02}} = \sqrt{5000} = 70.71 \text{ m/s} ] 6. Check: Units are m/s. Speed is reasonable (much less than speed of light).

What we did and why: We used the wave speed formula because tension and mass density were given. Always calculate ( \mu ) first if mass and length are provided.

Example 2 – Medium (Standing Waves)

Question: A string of length 1.5 m vibrates in its 3rd harmonic. If the wave speed is 120 m/s, what is the frequency of vibration?

Solution: 1. Given:
- ( L = 1.5 ) m
- ( n = 3 ) (3rd harmonic)
- ( v = 120 ) m/s 2. Find: ( f_3 = ? ) 3. Formula: ( f_n = \frac{n v}{2L} ) 4. Plug in:
[ f_3 = \frac{3 \times 120}{2 \times 1.5} = \frac{360}{3} = 120 \text{ Hz} ] 5. Check: Units are Hz. Frequency is reasonable (audible range).

What we did and why: We used the standing wave formula for the nth harmonic. The key was recognizing that the 3rd harmonic means ( n = 3 ).

Example 3 – Exam-Style (Sonometer)

Question: In a sonometer experiment, a string of length 1 m and mass 0.01 kg is under a tension of 25 N. If the string vibrates in its 2nd overtone, what is the frequency of the tuning fork used?

Solution: 1. Given:
- ( L = 1 ) m
- ( m = 0.01 ) kg
- ( T = 25 ) N
- 2nd overtone = 3rd harmonic (( n = 3 )) 2. Find: ( f = ? ) 3. Formula: ( f = \frac{n}{2L} \sqrt{T/\mu} ) 4. Calculate ( \mu ):
[ \mu = \frac{m}{L} = \frac{0.01}{1} = 0.01 \text{ kg/m} ] 5. Plug in:
[ f = \frac{3}{2 \times 1} \sqrt{\frac{25}{0.01}} = \frac{3}{2} \sqrt{2500} = \frac{3}{2} \times 50 = 75 \text{ Hz} ] 6. Check: Units are Hz. Frequency is reasonable.

What we did and why: We used the sonometer formula because the question involved resonance with a tuning fork. The 2nd overtone corresponds to ( n = 3 ), not ( n = 2 )—this is a common trap!

COMMON MISTAKES

1. Mistake: Confusing harmonic number with overtone number.
2. Why it happens: Students think 1st overtone = 1st harmonic.

3. Correct approach: 1st overtone = 2nd harmonic (( n = 2 )), 2nd overtone = 3rd harmonic (( n = 3 )).

4. Mistake: Forgetting to calculate ( \mu ) (linear mass density).

5. Why it happens: Students plug in mass directly into ( v = \sqrt{T/\mu} ).

6. Correct approach: ( \mu = m/L ). Always calculate ( \mu ) first.

7. Mistake: Ignoring phase changes in reflection.

8. Why it happens: Students assume all reflections are the same.

9. Correct approach: Fixed end = inverted (( \pi ) phase shift). Free end = upright (no shift).

10. Mistake: Using ( v = f \lambda ) without checking if ( v ) is constant.

11. Why it happens: Students assume wave speed is always ( f \lambda ), even when tension changes.

12. Correct approach: ( v = \sqrt{T/\mu} ) is the primary formula. ( v = f \lambda ) is secondary.

13. Mistake: Misidentifying nodes and antinodes.

14. Why it happens: Students confuse them with crests and troughs.
15. Correct approach: Nodes = zero displacement. Antinodes = maximum displacement.

EXAM TRAPS

1. Trap: Giving mass instead of linear mass density (( \mu )).
2. How to spot it: Question says "mass = 0.02 kg" but doesn’t mention length.

3. How to avoid it: Always calculate ( \mu = m/L ). If ( L ) is missing, it’s likely given elsewhere in the question.

4. Trap: Asking for wavelength but giving frequency.

5. How to spot it: Question says "frequency = 50 Hz" but asks for wavelength.

6. How to avoid it: Use ( v = f \lambda ) only if ( v ) is constant. Otherwise, find ( v ) first using ( \sqrt{T/\mu} ).

7. Trap: Partial reflection questions with impedance mismatch.

8. How to spot it: Question mentions two strings of different densities joined together.
9. How to avoid it: Use ( R = \left( \frac{Z_2 - Z_1}{Z_2 + Z_1} \right)^2 ), where ( Z = \sqrt{T \mu} ).

1-MINUTE RECAP

(Spoken naturally, as if to a student the night before the exam.)

"Alright, listen up. For wave speed on a string, remember ( v = \sqrt{T/\mu} ). Tension up? Speed up. Thicker string? Speed down. Standing waves? Nodes at ends for fixed strings, antinodes for free ends. Harmonics: 1st harmonic = fundamental, 2nd harmonic = 1st overtone. Sonometer? Use ( f = \frac{n}{2L} \sqrt{T/\mu} ). Watch out for phase changes—fixed end flips the wave, free end doesn’t. Common traps: Don’t confuse overtone with harmonic number, always calculate ( \mu ), and check if ( v ) is constant before using ( v = f \lambda ). That’s it. Now go crush that exam."