Calculate the amplitude for a SHM using the equation x = 3sin2pt + 4cos3pt

The correct option is 5.

To understand why this is the correct amplitude, let's analyze the given equation for the displacement of a particle:
$$x=3\sin \left(2\pi t\right)+4\cos \left(3\pi t\right)$$

Typically, when we combine two perpendicular or parallel simple harmonic motions (SHM) of the same frequency, the resultant amplitude can be found using vector addition or trigonometric identities. Here, the expression is represented in the form:
$$x=A_1\sin \left({\omega }_{1}t\right)+A_2\cos \left({\omega }_{2}t\right)$$
where the individual amplitudes are:
$$A_1=3$$
and
$$A_2=4$$

Since the sine and cosine components are out of phase by 90 degrees (or $\pi /2$ radians), we can calculate the resultant amplitude ($A$) of the combined motion using the Pythagorean relationship for orthogonal components:
$$A=\sqrt{{A_1}^2+{A_2}^2}$$

Substituting the values of $A_1$ and $A_2$ into the formula:
$$A=\sqrt{3^2+4^2}$$
$$A=\sqrt{9+16}$$
$$A=\sqrt{25}=5$$

Thus, the resultant amplitude of the motion is 5.