Question

The displacement x of a particle in motion is given in terms of time by x (x - 4) = 1 - 5 $\cos \omega t$

• A. The particle executes SHM
• B. The particle executes oscillatory motion which is not SHM
• C. motion of the particle is neither oscillator; nor simple harmonic
• D. The particle is not acted upon by a force when it is at x = 4

Answer 1

The correct option is A The particle executes SHM

$\begin{array}{c}x\left(x-u\right)=1-5\cos wt\\ x^2-4x=1-5\cos wt\\ x=\frac{4\pm \sqrt{16+4\left(1-5\cos wt\right)}}{2}\\ =\frac{4\pm \sqrt{20-2\cos wt}}{2}\\ =\frac{4\pm \sqrt{20\left(2{\sin }^2\right)wt}}{2}\\ =\frac{4\pm \sqrt{40}\sin wt}{2}\\ =2\pm \sqrt{10}\sin wt\\ So.particleperformSHM\\ Hence,theoptionAisthecorrectanswer.\end{array}$