Question

Two travelling waves y_(1)=4sin (3)/(2)(x-vt) cm and y_(2)=4sin (3)/(2)(x+vt) cm are superimposed.The distance between adjacent pressure antinodes is (in cm )​

Answer 1

Answer:

Given,

Y

1

=Asin[k(x+ct)] ...(i)

and Y

2

=Asin[k(x−ct)] .. .(ii)

By the principle of superposition, the resultant displacement of the particle is given by

Y=Y

1

+Y

2

Y=A[sin{k(x+ct)}+sin{k(x−ct)}]

By the formula

sinC+sinD=2sin

2

C+D

⋅cos

2

C−D

We have Y=2Asin

2

kx+kct+kx−kct

⋅cos

2

kx+kct−kx+kct

y=2Asinkx⋅coskct

For first antinode

sinkx

1

=1

sinkx

1

=sin

2

π

kx

1

=

2

π

...(iii)

For second antinode

sinkx

2

=−1

sinkx

2

=sin

2

3π

kx

2

=

2

3π

(iv)

∴ The distance between adjacent antinodes

kx

2

−kx

1

=

2

3π

−

2

π

k(x

2

−x

1

)=π

Δx=

k

π

Explanation:

you can try this method I hope it's helpfull for you