Question

Out of the following functions representing motion of a particle which represents $SHM$ ?
(A) $y=\sin \omega t-\cos \omega t$
(B) $y={\sin }^3\omega t$
(C) $y=5\cos (\frac{3\pi }{4}-3\omega t)$
(D) $y=1+\omega t+{\omega }^2{t}^2$

• A. Only $\left(A\right)$ and $\left(B\right)$
• B. Only $\left(A\right)$
• C. Only $\left(D\right)$ does not represent $SHM$
• D. Only $\left(A\right)$ and $\left(C\right)$

Answer 1

Answer: (D)

Only $\left(A\right)$ and $\left(C\right)$

(a) y = sinwt - coswt .

= √2 {$\frac{1}{\sqrt{2}}$ sinwt - $\frac{1}{\sqrt{2}}$coswt}

= √2{cos45°.sinwt - sin45°.coswt}

= √2sin(wt - 45°)

This is in the form of y = Asin(wt ± ∅)

So, this is the equation of SHM .

Period =$\frac{2\pi }{\omega }$

(b) sin³wt

Use formula of sin3x = 3sinx - 4sin³x

So, sin³wt = $\frac{1}{4}$ [ 3sinwt - sin3wt ]

Hence, you observed that this equation is a combination of two SHM. Hence, this is not SHM. But periodic motion.

Period = LCM of period { sinwt , sin3wt }

Period of sinwt = $\frac{2\pi }{\omega }$

period of sin3wt = $\frac{2\pi }{3\omega }$

so, period = $\frac{2\pi }{\omega }$

(c) 5cos($\frac{3\pi }{4}$ - 3wt)

= 5cos {-(3wt-$\frac{3\pi }{4}$)}

= 5cos(3wt - $\frac{3\pi }{4}$) [ cos(-∅) = cos∅]

Hence, this is SHM .

Period = $\frac{2\pi }{2\omega }$ = $\frac{\pi }{\omega }$

(d) x = 1 + wt + w²t²
At x →∞ , t→∞
There is no repetation of values . hence, this neither periodic nor SHM.