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 \(\frac{2\pi}{\omega}\).
2.	simple harmonic with period \(\frac{\pi}{\omega}\).
3.	periodic but not simple harmonic.
4.	non-periodic.

(a) 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,

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

=

\(-3(-\omega)sin(\frac{\pi}{4}-\omega t)\)

\(= 3\omega sin(\frac{\pi}{4}- \omega t)\)

Step 2: Find the acceleration of the particle.

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

= 3 ω \times (-

ω) cos(π/4-ωt)=-3ω²cos(π/4-ωt)

=-3ω²cos(π/4-ωt)

\Rightarrow a = -

ω²y

\Rightarrow

As acceleration, a

\propto - y

Hence, due to negative sign motion is SHM.

Step 3: Find the time period of the motion.

Clearly, from the equation,

ω' =

\Rightarrow \frac{2 π}{T'} =

ω⇒T'=2π/ω=2π/ω

So, the motion is SHM with period

\frac{2π}{ω}