The displacement of a particle is represented by the equation \(y= 3 \cos \left(\frac{\pi}{4}-\omega t \right)\). The motion of the particle is:
1.
simple harmonic with period \(\dfrac{2\pi}{\omega}\)
2.
simple harmonic with period \(\dfrac{\pi}{\omega}\)
3.
periodic but not simple harmonic
4.
non-periodic
Hint: In SHM, the acceleration is directly proportional to the displacement.
Step 1: Find the velocity of the particle.
Given, \(y = 3 \cos (\frac{\pi}{4}-\omega t)\)
The velocity of the particle is given by; \(v=\frac{dy}{dt}\) \(v= \frac{d}{dt}[3 \cos(\frac{\pi}{4}-\omega t)]\) \(v=-3(-\omega)\sin(\frac{\pi}{4}-\omega t)= 3\omega \sin(\frac{\pi}{4}- \omega t)\)
Step 2: Find the acceleration of the particle.
Acceleration, \(a = \frac{dv}{dt} \) \(a= \frac{d}{dt}[3\omega \sin(\frac{\pi}{4}-\omega t)]\) \(a=3\omega \cos\left(\frac{\pi}{4} - \omega t\right) \cdot (-\omega)= -3\omega^2 \cos\left(\frac{\pi}{4} - \omega t\right)\) \(\Rightarrow a=-\omega^2y\)
As acceleration, \(a\propto -y\)
Hence, due to the negative sign motion is SHM.
Step 3:Find the time period of the motion.
The time period of the particle in SHM is given by; \(T=\frac{2\pi}{\omega}\)
So, the motion is SHM with a period \(\frac{2\pi}{\omega}.\)
Hence, option (1) is the correct answer.
