Explanation:

We are given that

Displacement of a particle in S.H.M $\mathrm{x}(\mathrm{t})=3 \sin (20 \pi t)+4 \cos (20 \pi t)$

We have to find the amplitude of simple harmonic motion.

We know that general equation of displacement in SHM is given by $x(t)=c_{1} \cos (\omega t)+c_{1} \sin (\omega t)$

Then, amplitude $=A=\sqrt{c_{1}^{2}+c_{2}^{2}}$

By comparing with the given equation

Then, we get

$c_{1}=4, c_{2}=3$

Amplitude $=\sqrt{(3)^{2}+(4)^{2}}$

Amplitude $=\sqrt{25}=5$

It is always positive because it is maximum displacement from equilibrium position.

Hence, the amplitude of oscillation=5 $\mathrm{cm}$