Ncert Exercises & Solutions

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{π}{4} - ω t)

$y = 3 \cos (\frac{\pi}{4}-\omega t)$

The velocity of the particle is given by;

v = \frac{d y}{d t}

$v=\frac{dy}{dt}$

v = \frac{d}{d t} [3 \cos (\frac{π}{4} - ω t)]

$v= \frac{d}{dt}[3 \cos(\frac{\pi}{4}-\omega t)]$

v = - 3 (- ω) \sin (\frac{π}{4} - ω t) = 3 ω \sin (\frac{π}{4} - ω 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{d v}{d t}

$a = \frac{dv}{dt}$

a = \frac{d}{d t} [3 ω \sin (\frac{π}{4} - ω t)]

$a= \frac{d}{dt}[3\omega \sin(\frac{\pi}{4}-\omega t)]$

a = 3 ω \cos (\frac{π}{4} - ω t) \cdot (- ω) = - 3 ω^{2} \cos (\frac{π}{4} - ω 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 = - ω^{2} y

$\Rightarrow a=-\omega^2y$
As acceleration,

a \propto - y

$a\propto -y$
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 π}{ω}

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

So, the motion is SHM with a period

\frac{2 π}{ω} .

$\frac{2\pi}{\omega}.$
Hence, option (1) is the correct answer.