The motion of a particle is described by the equation $a = -bx$,where $a$ is the acceleration,$x$ is the displacement from the equilibrium position,and $b$ is a constant. The periodic time will be

The figure shows the circular motion of a particle. The radius of the circle is $B$,the period is $30 \ s$,the sense of revolution is clockwise,and the initial position at $t = 0$ is at an angle of $45^\circ$ (or $\pi/4$ radians) with the positive $y$-axis in the first quadrant. The simple harmonic motion of the $x$-projection of the radius vector of the rotating particle $P$ is:

The periodic time of $SHM$ is also measured by $T = 2 \pi \sqrt{\frac{\text{displacement}}{\text{acceleration}}}$. Can we say that the time period depends on the displacement?

What is a linear harmonic oscillator? And what is a non-linear oscillator?

Write the equation representing the relation between time period and frequency.

The displacement of an oscillator is given by $x = a \sin \omega t + b \cos \omega t$,where $a, b$ and $\omega$ are constants. Then: