Question 49: Oscillations

Question

For particle P revolving round the centre O with radius of circular path $$\mathrm{r}$$ and angular velocity $$\omega$$, as shown in below figure, the projection of OP on the $$x$$-axis at time $$t$$ is

Accepted Answer

After time $t$, the angular displacement will be
$	heta=\omega t$
Total angular displacement from $x$-axis.
$$
	heta_{	ext {total }}=\omega t+\frac{\pi}{6} \quad\quad\left(\because 30^{\circ}=\frac{\pi}{6}ight)
$$
Now, OP has two component
The horizontal component will be the projection along $x$-axis
$$
=r \cos \left(	heta_{	ext {Total }}ight)=r \cos \left(\omega t+\frac{\pi}{6}ight)
$$

Answer

$$x(t)=\operatorname{cos}\left(\omega t-\frac{\pi}{6} \omega\right)$$

Answer

$$x(t)=\operatorname{cos}(\omega t)$$

Answer

$$x(t)=r \cos \left(\omega t+\frac{\pi}{6}\right)$$

Answer

$$x(t)=r \sin \left(\omega t+\frac{\pi}{6}\right)$$

Oscillations

For particle P revolving round the centre O with radius of circular path $\mathrm{r} and angular velocity \omega, as shown in below figure, the projection of OP on the x-axis at time t$ is

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