The correct option is **D** Antinodes are formed at x values given by π6, π2, 5π6, 7π6,…

Given that,

y1=4sin(3x−2t)cm

y2=4sin(3x+2t)cm

By using superposition of waves, the net displacement

y=y1+y2

⇒y=4sin(3x−2t)+4sin(3x+2t)

⇒y=4[sin(3x−2t)+sin(3x+2t)]

Applying: sin(A−B)+sin(A+B)=2sinAcosB

⇒y=4(2sin3xcos2t)

⇒y=8sin3xcos2t

Now for maximum displacement, considering maximum value of cos2t=1, we have:

Amplitude, A=8sin(3x)

The displacement at x=2.3 cm is

A′=8sin[3(2.3)]

⇒A′=8sin(6.9)

⇒A′=4.63 cm

Nodes are formed where, amplitude or, intensity =0 i.e.,

IR=A2=0

⇒A2=sin2(3x)=0

⇒sin(3x)=0

3x=0, π, 2π, 3π

∴x=0, π3, 2π3, π, 4π3…

Anti-node is always formed between two nodes in a standing waves.

The position of antinodes are

π6, π2, 5π6,7π6…

So, we see corresponding location of anti-nodes lie between the calculated node locations.

Hence, option (a), (c), (d) are correct.

The correct option is **D** Antinodes are formed at x values given by π6, π2, 5π6, 7π6,…

Given that,

y1=4sin(3x−2t)cm

y2=4sin(3x+2t)cm

By using superposition of waves, the net displacement

y=y1+y2

⇒y=4sin(3x−2t)+4sin(3x+2t)

⇒y=4[sin(3x−2t)+sin(3x+2t)]

Applying: sin(A−B)+sin(A+B)=2sinAcosB

⇒y=4(2sin3xcos2t)

⇒y=8sin3xcos2t

Now for maximum displacement, considering maximum value of cos2t=1, we have:

Amplitude, A=8sin(3x)

The displacement at x=2.3 cm is

A′=8sin[3(2.3)]

⇒A′=8sin(6.9)

⇒A′=4.63 cm

Nodes are formed where, amplitude or, intensity =0 i.e.,

IR=A2=0

⇒A2=sin2(3x)=0

⇒sin(3x)=0

3x=0, π, 2π, 3π

∴x=0, π3, 2π3, π, 4π3…

Anti-node is always formed between two nodes in a standing waves.

The position of antinodes are

π6, π2, 5π6,7π6…

So, we see corresponding location of anti-nodes lie between the calculated node locations.

Hence, option (a), (c), (d) are correct.