Question

The displacement of a particle is represented by the equation $y=3\text{cos}(\frac{\pi }{4}-2\omega t)$. The motion of the particle is

• A. Simple harmonic with period $\frac{2\pi }{\omega }$
• B. Simple harmonic with period $\frac{\pi }{\omega }$
• C. Periodic but not simple harmonic
• D. Non-periodic

Answer 1

The correct option is B Simple harmonic with period $\frac{\pi }{\omega }$

Given, $x=3\text{cos}(\frac{\pi }{4}-2\omega t)$
Velocity , $v=\frac{dy}{dt}=3\times 2\omega \text{sin}(\frac{\pi }{4}-2\omega t)$
Acceleration, $a=\frac{dv}{dt}=-12{\omega }^2\text{cos}(\frac{\pi }{4}-2\omega t)$
$\Rightarrow a=-4{\omega }^{2}x$
As, $a\propto x$ and negative sign shows that it is directed towards equilibrium (or mean position), hence particle will execute SHM.
$x=3\text{cos}(\frac{\pi }{4}-2\omega t)=3\text{cos}(2\omega t-\frac{\pi }{4})$
Comparing above with equation $x=r\text{cos}({\omega }^{\prime }t+\varphi )$, we have
${\omega }^{\prime }=2\omega$
$\frac{2\pi }{T^{\prime }}=2\omega$
$\Rightarrow T^{\prime }=\frac{\pi }{\omega }$