# Nootan Solutions Superposition of Wave-1 ISC Physics Class-11 Ch-27

Nootan Solutions Superposition of Wave-1 ISC Physics Class-11 Ch-27 Vol-2 Interference and Beats Nageen Prakashan. Step by step Solutions of Numericals of latest edition of Kumar and Mittal ISC Physics Part-2 Class-11 Nageen Prakashan Questions. Visit official Website **CISCE** for detail information about ISC Board Class-11 Physics.

## Nootan Solutions Superposition of Wave-1 ISC Physics Class-11 Ch-27

Class: | 11 |

Subject: | Physics Part-2 |

Chapter: 27 | Superposition of Wave-1 Interference and Beats |

Board | ISC |

Writer / Publications | Nootan / Nageen Prakashan / Kumar and Mittal |

Topics | Solved Numericals of page 972, 973 |

**Nootan Solutions Superposition of Wave-1 ISC Physics Class-11 Ch-27 Vol-2 Kumar and Mittal**

**Principle of Superposition of Waves :-**

Considering two waves, travelling simultaneously along the same stretched string in opposite directions as shown in the figure above. We can see images of waveforms in the string at each instant of time.

It is observed that the net displacement of any element of the string at a given time is the algebraic sum of the displacements due to each wave.

Let us say two waves are travelling alone and the displacements of any element of these two waves can be represented by y_{1}(x, t) and y_{2}(x, t). When these two waves overlap, the resultant displacement can be given as y(x,t).

**Mathematically, **

y (x, t) = y_{1}(x, t) + y_{2}(x, t)

As per the principle of superposition, we can add the overlapped waves algebraically to produce a resultant wave.

Let us say the wave functions of the moving waves are

y_{1} = f1(x–vt),

y_{2} = f2(x–vt)

……….

y_{n} = f_{n} (x–vt)

then the wave function describing the disturbance in the medium can be described as

y = f_{1}(x – vt)+ f_{2}(x – vt)+ …+ f_{n}(x – vt)

or,

y=∑ i=1 to n = fi (x−vt)

Consider a wave travelling along a stretched string given by, y_{1}(x, t) = A sin (kx – ωt) and another wave, shifted from the first by a phase φ, given as y_{2}(x, t) = A sin (kx – ωt + φ)

From the equations we can see that both the waves have the same angular frequency, same angular wave number k, hence the same wavelength and the same amplitude A.

Now, applying the superposition principle, the resultant wave is the algebraic sum of the two constituent waves and has displacement

y(x, t) = A sin (kx – ωt) + A sin (kx – ωt + φ)

As, sin A = sin B = 2sin (A+B)/2 . cos (A−B)/2

**The above equation can be written as,**

y(x, t) = [2A cos 1/2 ϕ] sin (kx − wt + 1/2ϕ)

The resultant wave is a sinusoidal wave, travelling in the positive X direction, where the phase angle is half of the phase difference of the individual waves and the amplitude as [2cos 1/2ϕ] times the amplitudes of the original waves.

**What is Interference of Light :-**

The phenomena of formation of maximum intensity at some points and minimum intensity at some other point when two (or) more waves of equal frequency having constant phase difference arrive at a point simultaneously, superimpose with each other is known as interference.

Nootan Solutions Superposition of Wave-1 ISC Physics Class-11 Ch-27

**(Page – 972, 973)**

**Question 1:**

**Two wave of the same frequency and same amplitude ……………………………….. the resultant wave be:**

**(i)………………**

**(ii)………………**

**(iii)………………**

**(iv)………………**

**Question 2:**

**Two sound wave emerging from ……………………. of the source.**

**Question 3:**

**…………………….**

**……………………**

**……………………**

**Question 16:**

**The wavelength of two ………………………. 332 ms-1.**

**Question 17:**

**When, at 15 …………………………… per degree C.**

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