Question

The displacement of a particle varies according to the relation $x=3sin100t+8co{s}^{2}50t$. Which of the following is/are correct about this motion.

• A. the motion of the particle is not S.H.M
• B. the amplitude of the S.H.M of the particle is 5 units
• C. the amplitude of the resultant S.H.M is $\sqrt{7}3$ units
• D. the maximum displacement of the particle from the origin is 9 units

Answer 1

Answer: (B), (D)

The correct options are
B the amplitude of the S.H.M of the particle is 5 units
D the maximum displacement of the particle from the origin is 9 units

$x=3\sin 100t+8{\cos }^{2}50t=3\sin 100t+4(1+\cos 100t)$ (using$2{\cos }^{2}A=1+\cos 2A$)
$x=3\sin 100t+4\cos 100t+4$
$\frac{dx}{dt}=300\cos 100t-400\sin 100t$
$\frac{d^2(x-4)}{d{t}^2}=-30000\sin 100t+40000\cos 100t=-100(3\sin 100t+4\cos 100t)=-100(x-4)$
so it is a SHM.
Amplitude of the motion,$A=\sqrt{3^2+4^2}=5units$
for max distance, $\frac{dx}{dt}=300\cos 100t-400\sin 100t=0\Rightarrow \tan 100t=\frac{3}{4}$
$\therefore \sin 100t=\frac{3}{5},\cos 100t=\frac{4}{5}$
now, $x_{max}=3\times \frac{3}{5}+4\times \frac{4}{5}+4=5+4=9units$
Ans:(B),(D)