Superposition of Waves Numerical on Interference Class-11 Nootan ISC Physics Ch-27. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Superposition of Waves Numerical on Interference Class-11 Nootan ISC Physics Ch-27
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Book | Nootan |
| Chapter-27 | Superposition of Waves-1 : Interference and Beats |
| Topics | Numericals on Interference of Waves |
| Academic Session | 2025-2026 |
Numericals on Interference of Waves
Superposition of Waves-1 : Interference and Beats Class-11 Nootan ISC Physics Ch-27 solutions of Kumar and Mittal
Q-1: Two waves of the same frequency and same amplitude a are reaching a point simultaneously. What should be the phase difference between the waves so that the amplitude of the resultant wave be : (i) 2a, (ii) √2a, (iii) a and (iv) zero?
Ans- we know that R² = a1² + a2² + 2 a1 a2 cos Φ
(i) R= 2a a1 = a2 = a
=> (2a)² = a² + a² + 2 a a cos Φ
=> 4a² = 2a² + 2a² cos Φ
=> 2a² cos Φ = 2a² / 2a² = 1 = cos 0°
=> Φ = 0°
(ii) R = √2 a
=> (√2 a)² = a² + a² + 2 a a cos Φ
=> (2a)² = 2a² + 2a² cos Φ
=> 2a² cos Φ = 2a² – 2a² = 0 = cos 90°
=> Φ = 90°
(iii) R = a then
=> a² = a² + a² + 2 a a cos Φ
=> a² = 2a² + 2a² cos Φ
=> 2a² cos Φ = a² – 2a²
=> 2a² cos Φ = a²
=> cos Φ = -1 = cos 120°
=> Φ = 120°
(iv) R = 0 then
=> 0² = a² + a² + 2 a a cos Φ
=> 0 = 2a² + 2a² cos Φ
=> cos Φ = -1 = cos 180°
=> Φ = 180°
Q-2: Two sound waves emerging from a source reach a point simultaneously along two paths. When the path difference is 12 cm or 36 cm, then there is silence at the point. If the speed of sound in air be 330 m s^-1, then calculate the frequency of the source.
Ans-2 Suppose that the frequency of sound is n.
Path Difference = 36 – 12 = 24 cm = 24 / 100 m
Since, v = n λ
=> n = v / λ = 330 / (24/100)
=> 55 x 25 = 1357 Hz
Q-3: In the experiment with Quincke’s tube, the tube is displaced through a distance of 30 cm to obtain two successive maximum sound. If the velocity of sound is 330 m s^-1, find the wavelength and frequency of the sound.
Ans-3 Suppose that the wavelength & frequency of sound is λ and n respectively
Given, λ / 2 = 30
=> λ = 2 x 30 = 60 cm
n = v / λ = 330 / (60/100)
=> 5.5 x 100 = 550 Hz
Q-4: The minimum distance through which one tube, in a Quincke’s tube, is to be shifted within the other, in order to get two such positions when no sound is heard, is 16.6 cm. Determine the frequency of the source of sound. (Speed of sound in air = 332 m s^-1.)
Ans-4 Suppose that the frequency is n then
Given λ / 2 = 16.6 cm
=> λ = 2 x 16.6 = 33.2 cm
n = v / λ = 332 x 100 / 33.2
=> 332000 / 332 = 1000 Hz
— : End of Superposition of Waves Numerical on Interference Class-11 Nootan ISC Physics Ch-27. :–
Return to : – – Nootan Solutions for ISC Physics Class-11
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