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🌊 WAVES (Mechanical Waves)

Stage 3 – Page 2 : Standing Waves on a Stretched String

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1️⃣ Fixed End & Free End – Boundary Conditions

📌 Fixed End

• Displacement = 0
• Always a Node

📌 Free End

• Slope = 0
• Always an Antinode

⚠ Boundary conditions decide allowed wavelengths.

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2️⃣ String Fixed at Both Ends

Let length of string = L

Allowed Wavelengths

λₙ = 2L / n   (n = 1, 2, 3, …)

Allowed Frequencies

fₙ = n (v / 2L)

Where wave speed:

v = √(T / μ)

• T = tension in string
• μ = mass per unit length

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3️⃣ Harmonics on Fixed–Fixed String

• n = 1 → Fundamental (1st harmonic)
• n = 2 → 2nd harmonic
• n = 3 → 3rd harmonic
• All harmonics are present

✔ Fundamental frequency is the lowest possible frequency.

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4️⃣ String Fixed at One End & Free at Other End

Length of string = L

Allowed Wavelengths

λₙ = 4L / (2n − 1)

Allowed Frequencies

fₙ = (2n − 1) (v / 4L)

• n = 1 → Fundamental
• n = 2 → 3rd harmonic
• n = 3 → 5th harmonic

⚠ Even harmonics are absent.

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5️⃣ Comparison Table (Very Important)

Feature	Fixed–Fixed String	Fixed–Free String
Ends	Node–Node	Node–Antinode
Harmonics	All present	Only odd
Fundamental λ	2L	4L
Fundamental f	v / 2L	v / 4L

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6️⃣ Effect of Physical Parameters

• Increase tension → frequency increases
• Increase mass density → frequency decreases
• Frequency ∝ √T
• Frequency ∝ 1/√μ

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7️⃣ JEE Advanced Concept Traps

• Mixing fixed–free with fixed–fixed formulas
• Wrong harmonic counting
• Forgetting √T relation
• Using λ instead of L directly

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🎯 Final Takeaway

✔ Boundary conditions decide wave pattern ✔ Harmonics differ for different end conditions ✔ Formulas must be chosen carefully

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Stage 3 – Page 2 Completed Successfully ✅