First slide

Simple hormonic motion

Question

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

Moderate

A

simple harmonic with particle is $\frac{{2\pi }}{\omega }$
B

simple harmonic with period π/ω
C

periodic but not simple harmonic
D

non periodic

Solution

$y = 3\cos \left( {\frac{\pi }{4} - 2\omega t} \right)$
$\Rightarrow y = 3\cos \left( {2\omega t - \frac{\pi }{4}} \right)$
Acceleration of the particle is $\frac{{{d^2}y}}{{d{t^2}}}$
Now $\frac{{{d^2}y}}{{d{t^2}}} = - {\left( {2\omega } \right)^2} \times 3\cos \left( {2\omega t - \frac{\pi }{4}} \right)$
$\Rightarrow \frac{{{d^2}y}}{{d{t^2}}} = - 4{\omega ^2}y$
$\therefore \frac{{{d^2}y}}{{d{t^2}}} \propto \left( { - y} \right)$
The above equation represents SHM. Whose amplitude is 3 unit, time period is $T = \frac{{2\pi }}{{2\omega }} = \frac{\pi }{\omega }$ and initial phase is - π/4.

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