Q4. A 384 Hz fork is used and the speed of sound is 345 m s. How far apart should the first two resonance lengths be?
l2 − l1 = λ/2 = 345/(2 × 384), about 0.449 m, that is about 44.9 cm.
Q5. The first two resonance lengths are 25.4 cm and 78.0 cm. Find the end correction.
e = (l2 − 3l1)/2 = (78.0 − 76.2)/2 = 0.9 cm.
Q6. Why is the second resonance fainter than the first?
At the second resonance the sound travels down a longer column and back, so more of its energy is lost on the way, mostly by friction and heat exchange at the walls, and the response is weaker. It may therefore be harder to identify by ear. Loudness and the width of the resonance peak also depend on coupling and losses; a longer column alone does not prove a particular change in sharpness.
Q7. NCERT Exercise 14.15 (page 299): a metre-long tube with a movable piston resonates with a 340 Hz fork at 25.5 cm and 79.3 cm. Neglecting end effects, estimate the speed of sound.
Neglecting end effects, the first resonance is a quarter wavelength: λ = 4 × 0.255 = 1.02 m, so v = 340 × 1.02 ≈ 347 m s (NCERT Exercise 14.15, page 299; NCERT’s answer, page 306); give that in an exam. Strictly, the readings are not quite consistent with zero end correction, since 79.3 cm is not 3 × 25.5 = 76.5 cm. The difference method, v = 2 × 340 × (0.793 − 0.255) ≈ 366 m s, removes the end correction, which these readings put at about 1.4 cm (reasoned).