CBSE Class 11 Physics Wave Motion Assignment

CBSE CLASS 12 PHYSICS
WAVE MOTION
MISCELLANEOUS ASSIGNMENT

MISCELLANEOUS ASSIGNMENT

Comprehension-1

Doppler's effect is of great importance in determining the velocity of approach or recession of heavenly bodies towards or away from the earth. When the light coming from a distant star is examined by a spectroscope, the spectrum consists of well defined lines. By Doppler's effect we know that when a source of light moves towards the observer, the frequency of light increases (i.e., the wavelength decreases) and vice versa. When the star approaches the earth, the lines are shifted towards the shorter wavelength i.e., towards the blue end of the spectrum. On the other hand, when the star moves away from the earth, the lines are shifted towards the red end. By measuring the shift, the velocity of the star can be calculated.
Let a star be moving with a velocity v with respect to the earth. Now by Doppler's principle, the altered frequency n' is given by
n' = \( \frac{cn}{c + v} \)
where c is the velocity of light and n is the true frequency of the light waves.
If λ' and λ are the altered and true wavelengths respectively, then
the change in wavelength dλ = λ \( \frac{v}{c} \).
Thus knowing the change in wavelength, v can be calculated.

1. A radar operates at wavelength 50 cm. If the beat frequency between transmitted signal and signal reflected from an aircraft is equal to 1 kHz at the radar location, the velocity of approaching aircraft will be
(a) 900 m/sec
(b) 450 m/sec
(c) 1000 m/sec
(d) 500 m/sec

Answer: 250 m/sec (none of the printed options matches)

1. The echo from an approaching aircraft is Doppler shifted twice: once when the aircraft receives the wave and once when it reflects it. So the beat frequency is Δf = 2v/λ.
2. v = Δf × λ / 2 = 1000 × 0.50 / 2 = 250 m/s.
3. This value is not among options (a) to (d), so the printed key (b) 450 m/sec cannot be justified; the correct velocity is 250 m/s.

Teacher's Note:
a) The factor 2 comes from the reflection; a one-way Doppler shift would give 500 m/s, which is option (d).
b) Always convert the wavelength to metres (50 cm = 0.5 m) before using the formula.

2. A radio wave of frequency 840 MHz is sent towards an aeroplane. The frequency of the radio echo has a frequency 2.8 kHz more than the original frequency. The velocity of the aeroplane is
(a) 3 km s^-1
(b) 2 km s^-1
(c) 4 km s^-1
(d) 0.1 km s^-1

Answer: 0.5 km s^-1 (none of the printed options matches)

1. For reflection from a moving target, Δf/f = 2v/c, so v = c Δf / (2f).
2. v = \( \frac{3 \times 10^8 \times 2.8 \times 10^3}{2 \times 840 \times 10^6} \) = 500 m/s = 0.5 km/s.
3. This value is not among the options, so the printed key (d) 0.1 km s^-1 is wrong; the correct velocity is 0.5 km s^-1.

Teacher's Note:
a) An aeroplane moving at 0.5 km/s (1800 km/h) is a sensible speed for a jet, whereas 2 to 4 km/s would be unrealistic.
b) Use c = 3 × 10⁸ m/s for radio waves.

3. A star is moving away from the Earth with a velocity of 10⁵ ms^-1. The shift in the spectral line of wavelength 5700 Å as observed on Earth is
(a) 0.53 Å
(b) 1.06 Å
(c) 1.90 Å
(d) 3.08 Å

Answer: (c) 1.90 Å

1. From the passage, dλ = λ v / c.
2. dλ = 5700 × 10⁵ / (3 × 10⁸) = 1.9 Å.
3. The star recedes, so the line shifts towards the red end by 1.9 Å.
4. Hence, the correct option is (c).

Teacher's Note:
a) Light speed c = 3 × 10⁸ m/s is used; v/c is only 1/3000, so the shift is small.
b) The shift is a red shift because the source moves away.

4. Radio waves are sent towards aeroplane flying high. Change in frequency will be
(a) higher when plane is moving away with speed v compare to change in frequency when plane is moving towards earth with same speed
(b) higher when plane is moving towards earth with speed v compare to change in frequency when plane is moving away with same speed.
(c) change in frequency is independent of whether plane moving towards or away.
(d) none of these

Answer: (c) change in frequency is independent of whether plane moving towards or away.

1. For an approaching plane the echo frequency rises by about 2vf/c, and for a receding plane it falls by about 2vf/c.
2. The size of the change is the same (for v much less than c); only the sign differs.
3. Hence, the correct option is (c).

Teacher's Note:
a) The shifts are equal only to a first approximation, since the exact expressions use (c - v) and (c + v).
b) Moving towards gives an increase, moving away gives a decrease.

Comprehension-2

Any function y = f (ax ± bt) defines wave provided function is bounded.
wave velocity = b/a
frequency is b/2π
Amplitude is defined as maximum displacement of particle

5. The equation of transverse wave in a string is y = 8 sin [1.256 x - 25.12 t] cm, x is in cm.
What is the amplitude of the wave?
(a) 8 cm
(b) 1.256 cm
(c) 25.12 cm
(d) 0.0208 m

Answer: (a) 8 cm

1. Compare with y = A sin (kx - ωt): A = 8 cm, k = 1.256 cm^-1, ω = 25.12 rad/s.
2. The amplitude is the coefficient of the sine, 8 cm.
3. Hence, the correct option is (a).

Teacher's Note:
a) The numbers 1.256 and 25.12 are k and ω, not the amplitude.
b) The answer is in cm because y is given in cm.

6. The frequency of the wave is
(a) 1 Hz
(b) 2 Hz
(c) 3 Hz
(d) 4 Hz

Answer: (d) 4 Hz

1. ω = 25.12 rad/s, so by the passage the frequency is b/2π.
2. f = 25.12 / (2 × 3.14) = 4 Hz.
3. Hence, the correct option is (d).

Teacher's Note:
a) 2π × 4 = 25.12, which confirms the answer.
b) Frequency in hertz = angular frequency divided by 2π.

7. λ is
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm

Answer: (d) 5 cm

1. The wave number is k = 1.256 cm^-1 = 2π/λ.
2. λ = 2π/k = 6.28 / 1.256 = 5 cm.
3. Hence, the correct option is (d).

Teacher's Note:
a) Since 1.256 = 2π/5, the wavelength is exactly 5 cm.
b) The coefficient of x gives the wave number, not the wavelength.

8. The speed of the wave is
(a) 5 cm s^-1
(b) 4 cm s^-1
(c) 20 cm s^-1
(d) 332 m s^-1

Answer: (c) 20 cm s^-1

1. Wave velocity = b/a = ω/k = 25.12 / 1.256 = 20 cm/s.
2. Cross-check: v = fλ = 4 × 5 = 20 cm/s.
3. Hence, the correct option is (c).

Teacher's Note:
a) The wave is slow because it is a wave on a string, not a sound wave; 332 m/s is the speed of sound in air.
b) Both the form b/a and fλ give the same value.

Each of the questions given below consists of two statements, an assertion (A) and reason (R). Select the number corresponding to the appropriate alternative as follows

(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

9. A: There is no effect of change in pressure on velocity of sound in air, provided temperature remains constant.
R: Because of change in pressure, density of air also changes. Therefore p/ρ becomes constant.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (a)

1. The speed of sound in air is v = √(γp/ρ), which depends on the ratio p/ρ.
2. At constant temperature, p/ρ = RT/M is constant, because a change in pressure changes the density in the same proportion (R is true).
3. So the speed of sound does not change with pressure alone, and A is true and is explained by R.
4. Hence, the correct option is (a).

Teacher's Note:
a) Velocity of sound in a gas depends on temperature (v ∝ √T), not on pressure.
b) This is why sound travels at the same speed on a mountain and at sea level at the same temperature.

10. A: Sound passes through air in the form of longitudinal waves.
R: Air possesses volume elasticity, which is the requirement for the propagation of longitudinal wave.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (a)

1. Sound in a gas is a series of compressions and rarefactions along the direction of travel, so it is longitudinal (A is true).
2. Longitudinal waves need a medium with volume (bulk) elasticity, and air has it (R is true).
3. R is the reason for A. Hence, the correct option is (a).

Teacher's Note:
a) Gases cannot sustain shear, so they cannot carry transverse waves.
b) The speed of a longitudinal wave in a gas involves the bulk modulus.

11. A: The apparent frequency is greater when a source of sound approaches a stationary listener than a listener approaches a stationary source with same velocity.
R: Doppler's effect is not observed when motion of source and listener is such that distance between them remains the same.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (b)

1. Source approaching: f₁ = f v/(v - u). Listener approaching: f₂ = f (v + u)/v.
2. f₁/f₂ = v²/(v² - u²), which is greater than 1, so A is true.
3. R is also true (no change in separation means no Doppler shift), but it does not explain why the two cases differ.
4. Hence, the correct option is (b).

Teacher's Note:
a) The two cases are not symmetrical because the medium (air) is at rest for only one of them.
b) For small u/v the two frequencies are almost equal.

12. A: Waves of length 50 cm are propagating on a string. The phase difference between 2 points separated by 5 cm is 0.2 π radian.
R: Phase difference = (2π/λ)x, where x is the path difference.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (a)

1. Phase difference = (2π/λ) x = (2π/50) × 5 = 0.2π rad, so R is true and A follows from it.
2. So A is true and R explains it.
3. Hence, the correct option is (a).

Teacher's Note:
a) Use the same unit (cm) for λ and x.
b) 0.2π rad is equal to 36°.

13. A: If the tension in the string is doubled, then frequency of fundamental node becomes two times.
R: According to law of tension n ∝ T, where T is tension in the string.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (d)

1. For a stretched string, the fundamental frequency is n = \( \frac{1}{2L}\sqrt{\frac{T}{\mu}} \), so n ∝ √T.
2. If T is doubled, the frequency becomes √2 times, not two times, so A is false.
3. The law of tension reads n ∝ √T (the root sign is lost in the printed R), so R, read with the root, is a true statement.
4. Hence, the correct option is (d).

Teacher's Note:
a) The printed key (b) is wrong: A is false, because the frequency changes by √2, not 2.
b) Frequency is proportional to the square root of tension, so it would double only when the tension becomes four times.

14. A: For an open organ pipe the first resonance position is at 40 cm. The third resonance position will be obtained at 200 cm.
R: In an open organ pipe n₃ = 5n₁.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (a)

1. Both statements describe an air column closed at one end, as in a resonance tube, where the first resonance is at λ/4 = 40 cm.
2. The resonances occur at λ/4, 3λ/4, 5λ/4, so the third resonance is at 5 × 40 = 200 cm (A is true).
3. The frequencies are n₁, 3n₁, 5n₁, so n₃ = 5n₁ (R is true), and R explains the 5 : 1 ratio of lengths.
4. Hence, the correct option is (a).

Teacher's Note:
a) If the pipe were really open at both ends, the resonances would be at 40, 80, 120 cm and n₃ = 3n₁; both statements would then be false and no option would fit.
b) The printed key (d) cannot be right in either reading, so the closed-pipe reading is used.

15. A: Antinodes are the points in a stationary wave where the displacement is maximum.
R: The distance between any two successive nodes or antinodes is λ/2.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (b)

1. At an antinode of a stationary wave the amplitude, and so the displacement, is maximum (A is true).
2. Successive nodes, or successive antinodes, are λ/2 apart (R is true).
3. R gives the spacing and does not explain why the displacement is maximum at an antinode.
4. Hence, the correct option is (b).

Teacher's Note:
a) The distance between a node and the next antinode is λ/4.
b) The explanation of A would be the superposition of two waves that add in phase at that point.

16. A: In closed end pipe only even harmonics are present.
R: In open end pipe only odd harmonic are present.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (d)

1. A closed pipe has a node at one end and an antinode at the other, so it supports only odd harmonics; A is false, since the even harmonics are absent.
2. An open pipe has antinodes at both ends and supports all harmonics, both odd and even.
3. As printed, R is also false, because an open pipe supports all harmonics, both odd and even. No option fits exactly, and the printed answer key gives (d).
4. Hence, the correct option is (d).

Teacher's Note:
a) As printed, both statements are false (a closed pipe gives odd harmonics only; an open pipe gives all harmonics). The printed key (d) is followed.
b) Remember: closed pipe gives odd harmonics only, open pipe gives all harmonics.

17. A: When a man is standing near a railway track, a fast-moving train may attract the man.
R: Due to larger gravitational force, because mass of moving train increases very much.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (c)

1. A fast moving train drags the air near it, so the speed of air is high between the man and the train and the pressure there is low (Bernoulli's principle).
2. The higher pressure on the other side pushes the man towards the train, so A is true.
3. The gravitational force is negligible, and the mass of the train does not increase noticeably at such speeds, so R is false.
4. Hence, the correct option is (c).

Teacher's Note:
a) The attraction is caused by the pressure difference of the air, not by gravity.
b) A mass increase appears only at speeds close to the speed of light.

18. A: The velocity of sound in air increases due to presence of moisture in it.
R: The presence of moisture in air lowers the density of air
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (a)

1. Water vapour (molar mass 18) replaces some air (molar mass about 29), so moist air has a lower density (R is true).
2. Since v = √(γp/ρ), a smaller density gives a larger speed of sound, so A is true.
3. R is the reason for A. Hence, the correct option is (a).

Teacher's Note:
a) The speed of sound is higher on humid days.
b) The pressure is the same, so only the density change matters.

19. A: If the number of beats produced by two sounding bodies is more than 10 per sec, then beats cannot be heard.
R: Velocity of sound is larger in solid bodies than in air.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (b)

1. The sensation of sound stays in the ear for about 0.1 s, so beats faster than about 10 per second cannot be distinguished (A is true).
2. The speed of sound in solids is greater than in air (R is true).
3. The speed of sound in a medium has no relation with the persistence of hearing, so R does not explain A.
4. Hence, the correct option is (b).

Teacher's Note:
a) Persistence of hearing (about 1/10 s) is the actual reason behind the limit of 10 beats per second.
b) Both statements are true, but they are unconnected.

20. A: If the wavelength of sound waves is less than 17 mm, then the human ear will not be able to detect the sound
R: Human ear is not able to detect ultrasonic waves.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (a)

1. In air (v = 340 m/s) a wavelength of 17 mm corresponds to f = 340/0.017 = 20000 Hz, which is the upper limit of audibility, so A is true.
2. Waves above 20 kHz are ultrasonic and the human ear cannot detect them (R is true).
3. R gives the reason for A. Hence, the correct option is (a).

Teacher's Note:
a) A shorter wavelength means a higher frequency, since v = fλ.
b) The audible range of the human ear is about 20 Hz to 20 kHz.