1. Write the given waves in comparable form

The two waves are:

$y_1 = 5\sin 2\pi(x-vt) \text{ cm}$ $y_2 = 3\sin 2\pi(x-vt+1.5) \text{ cm}$

Let

$\theta = 2\pi(x-vt)$

Then:

$y_1 = 5\sin\theta$

and

$y_2 = 3\sin(\theta + 2\pi\times 1.5)$

Since

$2\pi\times 1.5 = 3\pi,$

we get

$y_2 = 3\sin(\theta + 3\pi)$

Now,

$\sin(\theta+3\pi)= -\sin\theta$

because adding $\pi$ changes the sign, and $3\pi$ is an odd multiple of $\pi$.

So,

$y_2 = -3\sin\theta$

2. Add the two waves

The resultant displacement is:

$y = y_1 + y_2 = 5\sin\theta - 3\sin\theta = 2\sin\theta$

3. Find the amplitude of the resultant wave

The resultant wave is:

$y = 2\sin\theta$

Hence, the amplitude is:

$A = 2\text{ cm}$

4. Check with options

The correct option is:

A: 2 cm

5. Compare with stored correct answer

Stored correct answer: A

This matches our derived answer.