JEE Physics Waves MCQs Set 11

Answer: (d) 6

Explanation:
1. Closed pipe frequencies: 1500, 4500, 7500, ... = 1500(2m + 1).
2. 1500(2m + 1) ≤ 20 000 gives 2m + 1 ≤ 13, so the harmonics are 1, 3, 5, 7, 9, 11, 13.
3. That is 7 harmonics, i.e. fundamental plus 6 overtones.

Teacher's Note:
1. Count only odd harmonics.
2. Overtones = harmonics − 1 (the fundamental is not an overtone).

Answer: (c) y = −9 sin(bt − ax)

Explanation:
1. Energy ∝ amplitude², so the reflected amplitude is √0.81 × 10 = 9.
2. Reflection from a denser medium reverses the phase (π shift), and the reflected wave travels in the +x direction.
3. At x = 0 the reflected wave must equal −9 sin bt, which is y = −9 sin(bt − ax).

Teacher's Note:
1. Amplitude falls by √(fraction of energy).
2. A dense-medium reflection gives a phase change of π.

Answer: (b) 31.7 cm

Explanation:
1. For the fundamental mode, L = v/4f = 330/(4 × 260) = 0.317 m.
2. L = 31.7 cm.

Teacher's Note:
1. Closed-pipe fundamental: L = λ/4.
2. Convert metres to centimetres.

Answer: (a) n

Explanation:
1. Open tube of length L: n = v/2L.
2. After dipping half, the tube is closed with air column L/2: f = v/[4(L/2)] = v/2L = n.

Teacher's Note:
1. Dipping converts the pipe to a closed one of half the length.
2. The two changes cancel.

Answer: (a) closed at one end and 1 m long

Explanation:
1. The common difference is 170 Hz, and 425 : 595 : 765 = 5 : 7 : 9, which are odd harmonics, so the pipe is closed at one end.
2. Successive odd harmonics differ by 2f0, so f0 = 85 Hz.
3. L = v/4f0 = 340/(4 × 85) = 1 m.

Teacher's Note:
1. Only odd harmonics appear in a closed pipe.
2. Check that the ratios 5 : 7 : 9 hold.

Answer: (a) 45 cm

Explanation:
1. λ = 330/330 = 1 m. Air columns that resonate: λ/4 = 25 cm, 3λ/4 = 75 cm (the next, 125 cm, exceeds the tube).
2. Minimum water column corresponds to the longest air column: 120 − 75 = 45 cm.

Teacher's Note:
1. The tube is closed at the water end.
2. The longest resonant air column gives the shortest water column.

Answer: (d) 100 cm

Explanation:
1. Closed pipe first overtone: 3v/4L = 3v/(4 × 0.5) = 1.5v.
2. Open pipe second overtone: 3v/2L′ = 1.5v, so L′ = 1 m = 100 cm.

Teacher's Note:
1. Second overtone of an open pipe is the 3rd harmonic.
2. First overtone of a closed pipe is also the 3rd harmonic.

Answer: (c) 20 Hz

Explanation:
1. Consecutive harmonics of a closed pipe differ by 2f0: 500 − 460 = 40, so f0 = 20 Hz.
2. Check: 460 = 23 × 20 and 500 = 25 × 20.

Teacher's Note:
1. The difference of successive odd harmonics is twice the fundamental.
2. Verify that both frequencies are odd multiples.

Answer: (b) 339 m/s

Explanation:
1. Successive resonances differ by λ/2: λ/2 = 93 − 40 = 53 cm, so λ = 106 cm.
2. v = fλ = 320 × 1.06 = 339 m/s.

Teacher's Note:
1. The end correction cancels in the difference.
2. Convert cm to m.

Answer: (a) 120 cm

Explanation:
1. Smallest length is λ/4 = 40 cm.
2. Next resonance is at 3λ/4 = 3 × 40 = 120 cm.

Teacher's Note:
1. Closed-pipe lengths are in the ratio 1 : 3 : 5.
2. Next length = 3 × smallest.

Answer: (d) 20 m/s

Explanation:
1. The fixed end is a node, and the next node is λ/2 away, so λ/2 = 10 cm and λ = 20 cm.
2. v = fλ = 100 × 0.2 = 20 m/s.

Teacher's Note:
1. Adjacent nodes are λ/2 apart.
2. The fixed end itself is the first node.

Answer: (c) pluck at ℓ/4, touch at ℓ/2

Explanation:
1. In the second harmonic there are nodes at 0, ℓ/2 and ℓ, and antinodes at ℓ/4 and 3ℓ/4.
2. The string is plucked at an antinode (ℓ/4) and touched at a node (ℓ/2) to suppress the fundamental.

Teacher's Note:
1. Pluck at an antinode, touch at a node.
2. The fundamental has an antinode at ℓ/2, which must be touched.

Answer: (b) 508 Hz

Explanation:
1. The string was 508 Hz or 516 Hz. Increasing the tension raises its frequency (f ∝ √T).
2. If it were 516 Hz, the beats would increase. Beats decrease, so the string moved closer to 512 Hz: it was 508 Hz.

Teacher's Note:
1. More tension means higher frequency.
2. Beats falling means the frequencies are approaching.

Answer: (a) 0.02

Explanation:
1. f ∝ √T, so Δf/f = ½ΔT/T.
2. 6/600 = 0.01 = ½ΔT/T, so ΔT/T = 0.02.

Teacher's Note:
1. Differentiate f ∝ T^(1/2).
2. Beats = Δf.

Answer: (a) 254 Hz

Explanation:
1. The unknown is 254 Hz or 246 Hz.
2. Second harmonic: 508 Hz or 492 Hz. With 513 Hz the beats are 5 or 21.
3. Only 254 Hz gives 5 beats.

Teacher's Note:
1. Test each candidate against the second condition.
2. The second harmonic is twice the frequency.

Answer: (d) \( \frac1n=\frac1{n_1}+\frac1{n_2}+\frac1{n_3} \)

Explanation:
1. For the same tension and mass per unit length, f = (1/2ℓ)√(T/μ), so ℓ ∝ 1/f.
2. Total length ℓ = ℓ1 + ℓ2 + ℓ3, so 1/n = 1/n1 + 1/n2 + 1/n3.

Teacher's Note:
1. Length is inversely proportional to frequency.
2. Lengths add, not frequencies.

Answer: (c) 105 Hz

Explanation:
1. Adjacent harmonics differ by the fundamental: 420 − 315 = 105 Hz.
2. Check: 315 = 3 × 105 and 420 = 4 × 105.

Teacher's Note:
1. No frequency between them means they are consecutive harmonics.
2. The common difference is f0.

Answer: (a) the velocity of propagation of waves changes by 2%

Explanation:
1. v ∝ √T, so a 2% change in tension changes the velocity (and frequency) by 1%.
2. 15 Hz is 1% of the frequency, so the original frequency is 1500 Hz.
3. The wavelength of the fundamental is 2ℓ and is unchanged. The statement about a 2% change in velocity is therefore incorrect.

Teacher's Note:
1. Half the percentage change in T.
2. λ depends only on the string length.

Answer: (a) 1/2 m

Explanation:
1. k = 2π, so λ = 2π/k = 1 m.
2. For fixed ends the minimum length is λ/2 = 1/2 m.

Teacher's Note:
1. The shortest length corresponds to the fundamental mode.
2. Fixed ends are nodes, so ℓ = λ/2.

Answer: (b) y = −a cos(ωt + kx)

Explanation:
1. The other wave travels in the opposite direction with equal amplitude: ±a cos(ωt + kx).
2. At x = 0 the sum must vanish for all t: a cos ωt + y2(0) = 0, so y2(0) = −a cos ωt.
3. y2 = −a cos(ωt + kx), and the sum is 2a sin ωt sin kx.

Teacher's Note:
1. A node needs the two waves to cancel at x = 0.
2. Reflection from a fixed end changes phase by π.

Answer: (a) 10

Explanation:
1. f = (1/2ℓ)√(T/μ) and μ ∝ ρd².
2. f1/f2 = (ℓ2/ℓ1)√[(T1/T2)(ρ2d2²)/(ρ1d1²)] = (35/36)√[8 × 2/16] = 35/36.
3. The second wire has the higher frequency, 360 Hz, so the first has 350 Hz. Beats = 10.

Teacher's Note:
1. μ is proportional to density times diameter squared.
2. Beats = difference of the two frequencies.

Answer: (d) 5

Explanation:
1. f = (p/2ℓ)√(T/μ) with T = Mg, p = number of loops.
2. Same f and ℓ: 5√9 = 3√M, so √M = 5.

Teacher's Note:
1. Loops × √T is constant.
2. Fewer loops need more tension.

Answer: (c) 445 Hz

Explanation:
1. The string is 435 Hz or 445 Hz from the first test.
2. With 436 Hz, 445 Hz gives 9 beats, while 435 Hz would give 1 beat.
3. So the string is 445 Hz.

Teacher's Note:
1. Test both candidates against the second fork.
2. Only one candidate fits both.

Answer: (b) 3.6 N

Explanation:
1. ω = 240π, so f = 120 Hz. k = 4π/5 cm⁻¹, so λ = 2π/k = 2.5 cm.
2. v = fλ = 120 × 2.5 = 300 cm/s = 3 m/s.
3. μ = 4 g/cm = 0.4 kg/m, so T = μv² = 0.4 × 9 = 3.6 N.

Teacher's Note:
1. Convert to SI units before using T = μv².
2. The cos kx factor gives the wavelength.

Answer: (a) \( \sqrt{\frac32} \)

Explanation:
1. Tension at the bottom is 2mg and at the top 2mg + mg = 3mg.
2. v ∝ √T, and the frequency stays constant, so λ ∝ v.
3. λ2/λ1 = √(3mg/2mg) = √(3/2).

Teacher's Note:
1. Frequency of the pulse does not change.
2. The rope's own weight adds to the tension at the top.