String breaks at fundamental resonance due to maximum amplitude

Steps:

• First stationary wave forms at fundamental frequency f, with length L = λ/4 for driven-clamped string.
• Resonance at f maximizes transverse displacement amplitude for given driving force.
• Elastic string stretches more under large amplitude, increasing tension to breaking point.
• Beyond f, amplitude drops until next (odd) harmonic, preventing higher breaking risk.

Why B is correct:

• At fundamental f, maximum amplitude causes peak tension via elastic extension, per wave equation where A ∝ Q-factor at resonance.

Why the others are wrong:

• A: 1/2 f produces no standing wave, as below fundamental; insufficient amplitude for breaking.
• C: 2 f is even harmonic, absent in this boundary setup; no resonance, low amplitude.
• D: 3 f is higher odd harmonic with nodes reducing effective amplitude and tension.

Final answer: B