Question

The motion of a particle is given by x = A sin ωt + B cos ωt. The motion of the particle is
(a) not simple harmonic
(b) simple harmonic with amplitude A + B
(c) simple harmonic with amplitude (A + B)/2
(d) simple harmonic with amplitude √A2+B2.

Answer 1

The motion of a particle is given by $x=A \sin \omega t+B \cos \omega t$. The motion of the particle is  simple harmonic with amplitude $\sqrt{A 2+B 2}$.

Explanation:

$A 2+B 2$

$x=\mathrm{B} \cos \omega \mathrm{t}+\mathrm{A} \sin \omega \mathrm{t}$                                   …(1)  

Acceleration, $a=d t 2 x d 2$

$=-B \omega 2 \cos \omega t-A \omega 2 \sin \omega t$

$=-\omega 2 x$

To undergo simple harmonic motion, the acceleration  

$a=-\mathrm{kx}$.                                          …(2)

So, from both the equations, it is observed that the simple harmonic motion is given with amplitude,  $A=B 2.2+A 2.$

Answer 2

Answer:

Refer to the above attachment.

Thank you.