Question

The position of a particle w.r.t origin varies according to the relation $x=3\sin 100t+8{\cos }^{2}50t$. 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 S.H.M is $\sqrt{73}\text{units}$.
• D. The maximum displacment of the particle from the origin is 9 units.

Answer 1

Answer: (B), (D)

The maximum displacment of the particle from the origin is 9 units.

Given ,
$x=3\sin 100t+8{\cos }^{2}50t$
The above equation can also be written as,
$x=3\sin 100t+\frac{8[1+\cos 100t]}{2}$
$\Rightarrow x=4+3\sin 100t+4\cos 100t$

This is not an expression for SHM about the point $x=0$, but we can rewrite the above equation as
$x-4=3\sin 100t+4\cos 100t$
This is an expression for the superposition of two SHMs of different amplitudes but same frequency, having a phase difference of $\frac{\pi }{2}$ about $x=4$.
The resultant of superposition of the two SHMs is written as,
$(x-4)=5\sin (100t+\varphi )\{\text{where}\tan \varphi =\frac{4}{3}\}$
Comparing this with $x=x_0+A\sin (\omega t+\varphi )$
Amplitude $\left(A\right)=5\text{units}$
Mean position $\left(x_0\right)=4\text{units}$
Maximum displacement $\left(x_{max}\right)=x_0+A=9\text{units}$.
Thus, options (b) and (d) are correct answers.