JEE Physics Waves MCQs Set 09

Answer: (b) 3 cm

Explanation:
1. Compare with \( y=2A\sin kx\cos\omega t \): \( k=\frac\pi3\ \text{cm}^{-1} \).
2. \( \lambda=\frac{2\pi}{k}=6\ \text{cm} \).
3. Adjacent nodes are \( \frac\lambda2=3\ \text{cm} \) apart.

Teacher's Note:
1. Nodes are separated by half a wavelength.
2. k = 2π/λ.

Answer: (a) 30%

Explanation:
1. \( f=\frac{1}{2L}\sqrt{\frac T\mu} \).
2. For constant f, \( L\propto\sqrt T \): \( \frac{L'}{L}=\sqrt{1.69}=1.3 \).
3. The length must increase by 30%.

Teacher's Note:
1. The fundamental frequency is inversely proportional to the length.
2. Take the square root of the tension ratio.

Answer: (d) 15.7 cm

Explanation:
1. Wave number \( k=0.6\ \text{cm}^{-1} \), so \( \lambda=\frac{2\pi}{0.6}=10.5\ \text{cm} \).
2. For the third harmonic the string holds three loops: \( L=\frac{3\lambda}{2}=15.7\ \text{cm} \).

Teacher's Note:
1. The nth harmonic of a string fixed at both ends has n half-wavelengths.
2. Get λ from the wave number.

Answer: (d) 1 : 3

Explanation:
1. Same material, so the mass per unit length \( \mu\propto r^2 \): \( \mu_A=4\mu_B \) and \( \sqrt{\mu_A}=2\sqrt{\mu_B} \).
2. First overtone of A: \( f=\frac{2}{2L_A}\sqrt{\frac T{\mu_A}} \). Second overtone of B: \( f=\frac{3}{2L_B}\sqrt{\frac T{\mu_B}} \).
3. Equate: \( \frac{1}{L_A\cdot2\sqrt{\mu_B}}=\frac{3}{2L_B\sqrt{\mu_B}}\Rightarrow\frac{L_A}{L_B}=\frac13 \).

Teacher's Note:
1. The first overtone is the second harmonic and the second overtone is the third harmonic.
2. Mass per unit length is proportional to the area of cross-section.

Answer: (c) 0.5 m/s

Explanation:
1. The second profile is the first shifted by 1 m in the +x direction.
2. Shift 1 m in 2 s gives speed \( v=\frac12=0.5\ \text{m/s} \).

Teacher's Note:
1. A travelling wave y(x, t) = f(x − vt).
2. Compare x − 1 with x − v(2).

Answer: (b) 1000 Hz

Explanation:
1. Moving the piston from a maximum to the next minimum changes the path (twice the movement) by \( \frac\lambda2 \), so the piston moves \( \frac\lambda4 \).
2. \( \frac\lambda4=8.75\ \text{cm}\Rightarrow\lambda=35\ \text{cm} \).
3. \( n=\frac v\lambda=\frac{350}{0.35}=1000\ \text{Hz} \).

Teacher's Note:
1. The reflected wave travels the extra distance twice.
2. Maximum to minimum is a path difference of λ/2.

Answer: (a) \( 2(I_1+I_2) \)

Explanation:
1. \( I_{max}=(\sqrt{I_1}+\sqrt{I_2})^2 \) and \( I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2 \).
2. Add: \( I_{max}+I_{min}=2(I_1+I_2) \).

Teacher's Note:
1. Intensity is proportional to the square of the amplitude.
2. The cross terms cancel when maximum and minimum are added.

Answer: (a) \( 2\sqrt3 \) cm

Explanation:
1. The amplitude at position x is \( 4\sin\frac{\pi x}{15} \).
2. At x = 5 cm: \( 4\sin\frac\pi3=4\times\frac{\sqrt3}2=2\sqrt3\ \text{cm} \).

Teacher's Note:
1. For a standing wave, each point oscillates with its own amplitude.
2. The maximum displacement is the amplitude factor at that position.

Answer: (d) \( 2I_0 \)

Explanation:
1. Path difference \( \Delta=5-4=1\ \text{m}=\frac\lambda4 \).
2. Phase difference \( \phi=\frac{2\pi}{\lambda}\Delta=\frac\pi2 \).
3. \( I=I_0+I_0+2I_0\cos\frac\pi2=2I_0 \).

Teacher's Note:
1. Find S2D with Pythagoras' theorem first.
2. I = I1 + I2 + 2√(I1 I2) cos φ.

Answer: (c) \( 2A\cos\frac\phi4 \)

Explanation:
1. The phase difference is \( \delta=\frac\phi2 \).
2. Resultant amplitude \( =\sqrt{A^2+A^2+2A^2\cos\delta}=2A\cos\frac\delta2 \).
3. So the amplitude is \( 2A\cos\frac\phi4 \).

Teacher's Note:
1. Resultant amplitude of two equal waves: 2A cos(δ/2).
2. Here δ = φ/2.

Answer: (b) 120 Hz

Explanation:
1. Open pipe: \( f_o=\frac{v}{2L}=240\ \text{Hz} \).
2. Closed pipe: \( f_c=\frac{v}{4L}=\frac{f_o}{2}=120\ \text{Hz} \).

Teacher's Note:
1. A closed pipe has a wavelength twice as long for the fundamental.
2. Only odd harmonics are present in a closed pipe.

Answer: (a) 1 : 2

Explanation:
1. Closed pipe, first overtone (third harmonic): \( f=\frac{3v}{4L_1} \).
2. Open pipe, third harmonic: \( f=\frac{3v}{2L_2} \).
3. Equate: \( \frac{1}{4L_1}=\frac1{2L_2}\Rightarrow\frac{L_1}{L_2}=\frac12 \).

Teacher's Note:
1. A pipe open at one end is a closed pipe: odd harmonics only.
2. Its first overtone is the 3rd harmonic.

Answer: (b) 400 Hz

Explanation:
1. For a given string under constant tension, \( f\propto\frac1L \).
2. \( f_2=256\times\frac{25}{16}=400\ \text{Hz} \).

Teacher's Note:
1. A shorter vibrating length means a higher frequency.
2. Use the inverse proportionality.

Answer: (a) 0.025 m

Explanation:
1. Successive resonances differ by \( \frac\lambda2=0.35-0.1=0.25\ \text{m}\Rightarrow\lambda=0.5\ \text{m} \).
2. First resonance: \( L_1+e=\frac\lambda4=0.125 \).
3. \( e=0.125-0.1=0.025\ \text{m} \).

Teacher's Note:
1. Check that 0.35 + e is three quarters of the wavelength.
2. The first resonance is at a quarter of the wavelength, measured including the end correction.

Answer: (d) 47 cm

Explanation:
1. First resonance: \( \frac\lambda4=15+1=16\ \text{cm}\Rightarrow\lambda=64\ \text{cm} \).
2. Second resonance: \( \frac{3\lambda}{4}-e=48-1=47\ \text{cm} \).

Teacher's Note:
1. Include the end correction in the effective length.
2. The second resonance is at 3λ/4.

Answer: (c) 2f m/s

Explanation:
1. Difference between successive resonating lengths \( =\frac\lambda2=1\ \text{m}\Rightarrow\lambda=2\ \text{m} \).
2. \( v=f\lambda=2f\ \text{m/s} \).

Teacher's Note:
1. Successive resonances are half a wavelength apart.
2. v = fλ.

Answer: (b) \( \frac{2(n+1)}{2n+1} \)

Explanation:
1. Open pipe, nth overtone: \( f_o=\frac{(n+1)v}{2L} \).
2. Closed pipe, nth overtone: \( f_c=\frac{(2n+1)v}{4L} \).
3. Ratio \( \frac{f_o}{f_c}=\frac{2(n+1)}{2n+1} \).

Teacher's Note:
1. The nth overtone of an open pipe is the (n+1)th harmonic.
2. The nth overtone of a closed pipe is the (2n+1)th harmonic.

Answer: (a) 1.0 m

Explanation:
1. The successive frequencies differ by 170 Hz. The ratio 425 : 595 : 765 is 5 : 7 : 9, odd integers, so the pipe is closed at one end.
2. For a closed pipe the difference is \( 2f_0=170\Rightarrow f_0=85\ \text{Hz} \) (and 425 = 5 × 85).
3. \( L=\frac{v}{4f_0}=\frac{340}{340}=1.0\ \text{m} \).

Teacher's Note:
1. Odd-integer ratios of the frequencies identify a closed pipe.
2. The fundamental of a closed pipe is v/4L.

Answer: (d) 99 Hz

Explanation:
1. Let the last be f. First \( =2f \) (octave). The difference over 25 steps: \( 2f-f=25\times3=75\Rightarrow f=75\ \text{Hz} \).
2. First fork \( =150\ \text{Hz} \).
3. 18th fork: \( 150-17\times3=99\ \text{Hz} \).

Teacher's Note:
1. There are 25 gaps between 26 forks.
2. Each step down in the sequence reduces the frequency by 3 Hz.

Answer: (a) \( \nu_a\gt \nu_b\gt \nu_c \)

Explanation:
1. At A the source has a velocity component towards O, so the observed frequency is above the true frequency.
2. At B the velocity is perpendicular to the line to O, so the observed frequency equals the true frequency.
3. At C the source moves away from O, so the observed frequency is below the true frequency.
4. Hence \( \nu_a\gt \nu_b\gt \nu_c \).

Teacher's Note:
1. Only the component of the source velocity along the line to the observer matters.
2. Approach raises the frequency; recession lowers it.

Answer: (c) \( \frac{2v_b(v+v_m)}{v^2-v_b^2}f \)

Explanation:
1. Direct sound from the band (behind it, the motorist is approaching the sound as it comes back, while the band moves away): \( f_1=\frac{v+v_m}{v+v_b}f \).
2. Sound reflected from the wall: the wall receives \( f_w=\frac{v}{v-v_b}f \) and the motorist, moving towards the wall, hears \( f_2=\frac{v+v_m}{v-v_b}f \).
3. Beat frequency \( f_2-f_1=(v+v_m)f\left[\frac1{v-v_b}-\frac1{v+v_b}\right]=\frac{2v_b(v+v_m)}{v^2-v_b^2}f \).

Teacher's Note:
1. Count the two sounds separately: direct and reflected.
2. Beats are the difference of the two frequencies.