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**Introductory Exercise 14.1**

**Ques 1: Prove that the equation y = a sin wt does not satisfy the wave equation and hence it does not represent a wave.****Sol: **

Hence, the given equation does not represent a wave equation.**Ques 2: A wave pulse is described by**** , where a, b and c are positive constants. What is the speed of this wave?****Sol: **Speed of wave

= c/b**Ques 3: The displacement of a wave disturbance propagating in the positive x-direction is given by and where, x and y are in metre. The shape of the wave disturbance does not change during the propagation. What is the velocity of the wave?Sol: **At t = 0, y is maximum at x = 0.

At t = 2 s, y is maximum at x = 1 m.

Hence, in 2 s, wave has travelled 2m in positive x-direction.

= +0.5 m/s

Here, x and y are in metre and t in second. In which direction and with what velocity is the pulse propagating? What is the amplitude of pulse?

Sol:

Speed of wave =

Amplitude = maximum value of y

= 10/5 = 2m

Here x and y are in metre and t in second. What will be the wave function representing the pulse at time t, if the pulse is propagating along positive x-axis with speed 2 m/s?

Sol:

âˆ´ Coefficient of t = 2 (coefficient of x) = 2 x 1 = 2 SI units.

**Introductory Exercise 14.2**

**Ques 1: The equation of a travelling wave is,y(x, t ) 0.02 sin Find: (a) The wave velocity and(b) the particle velocity at x = 0.2 m and t = 0.3 s.Given cos Î¸ = -0.85 where Î¸ = 34 radSol: **(a) Wave velocity =

Since, coefficient of t and coefficient of x are of same signs. Hence, wave is travelling in negative x-direction.

or

v = -5 m/s

(b) Particle velocity

Substituting x = 0.2 and t = 0.3, we have V

= (2) (-0.85) =-1.7 m/s

Sol:

and maximum particle speed, (v

From these two expressions, we can see that,

(v

(a) the frequency

(b) the phase difference between points 2.5 cm apart

(c) how long it takes for the phase at a given posit ion to change by 60Â°

(d) the velocity of a particle at point P at the instant shown.

Sol:

â‡’ Î» = 4cm

(d) At P, particle is at mean position. So, v = maximum velocity

= 125.7 cm/s= 1.26 m/s

Further,

...(i)

Sign of v

Sign of dy/dx, slope of y - x graph is also positive.

Hence, from Eq. (i) particle velocity is negative. v

(a) Write a wave function describing the wave.

(b) Find the transverse displacement of a point at x = 0.25 m at time t = 0.15 s.

(c) How much time must elapse from the instant in part (b) until the point at x = 0.25 m has zero displacement?

Sol:

Since wave is travelling along + ve x-direction, at and kx should have opposite signs. Further at t = 0, x = 0 the string has zero displacement and moving upward (in positive direction).

Hence at x = 0, we should have Asinwt not - Asinwt. Therefore, the correct expression is

(b) Putting x = 0.2 m and t = 0.15 s in the above equation we have,

y= 0.0354 m = 3.54 cm

(c) In part (b), y = A/âˆš2

From to y = 0, time taken is

= 4.2 Ã— 10

**Introductory Exercise 14.3**

**Ques 1. Calculate the velocity of a transverse wave along a string of length 2 m and mass 0.06 kg under a tension of 500 N.Sol: **

**Introductory Exercise 14.4**

**Ques 1. Spherical waves are emitted from a 1.0 W source in an isotropic non-absorbing medium. What is the wave intensity 1.0 m from the source?Sol:**

Sol:

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