When will the motion of a simple pendulum be simple harmonic?
Consider the diagram of a simple pendulum.
The bob is displaced through an angle $\theta$ shown.
The restoring torque about the fixed point $O$ is
$$\tau=-m g \sin \theta$$
If $\theta$ is small angle in radians, then $\sin \theta \approx \theta$
$$\Rightarrow \quad \tau \approx-m g \theta \Rightarrow \tau \propto(-\theta)$$
Hence, motion of a simple pendulum is SHM for small angle of oscillations.
What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?
Let $x=A \sin \omega t$ is the displacement function of SHM.
Velocity,
$$\begin{aligned} v & =\frac{d x}{d t}=A \omega \cos \omega t \\ v_{\max } & =A \omega|\cos \omega t|_{\max } \\ & =A \omega \times 1=\omega A \quad \left[\because|\cos \omega t|_{\max }=1\right] \ldots \text { (i) } \end{aligned}$$
$$\begin{aligned} \text { Acceleration, } a=\frac{d v}{d t}=-\omega A \cdot \omega \sin \omega t & \\ & =-\omega^2 A \sin \omega t \\ \left|a_{\max }\right| & =\left|\left(-\omega^2 A\right)(+1)\right| \quad \left[\because(\sin \omega t)_{\max }=1\right] \\ \left|a_{\max }\right| & =\omega^2 A\quad \text{... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { From Eqs. (i) and (ii), we get }\\ &\begin{array}{ll} \Rightarrow & \frac{v_{\max }}{a_{\max }}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega} \\ \Rightarrow \quad & \frac{a_{\max }}{v_{\max }}=\omega \end{array} \end{aligned}$$
What is the ratio between the distance travelled by the oscillator in one time period and amplitude?
The diagram represents
the motion of a particle executing SHM between $A$ and $B$.
Total distance travelled while it goes from $A$ to $B$ and returns to $A$ is
$$\begin{aligned} & =A O+O B+B O+O A \\ & =A+A+A+A=4 A \quad [\because O A=A] \end{aligned}$$
Amplitude $=O A=A$
Hence, ratio of distance and amplitude $=\frac{4 A}{A}=4$
In figure, what will be the sign of the velocity of the point $P^{\prime}$, which is the projection of the velocity of the reference particle $P . P$ is moving in a circle of radius $R$ in anti-clockwise direction.
As the particle on reference circle moves in anti-iclockwise direction. The projection will move from $P$ 'to $O$ towards left.
Hence, in the position shown the velocity is directed from $P^{\prime} \rightarrow P^{\prime \prime}$ i.e., from right to left, hence sign is negative.
Show that for a particle executing SHM, velocity and displacement have a phase difference of $\pi / 2$.
Let us assume the displacement function of SHM
$$\begin{aligned} \text{where,}\quad x & =a \cos \omega t \\ a & =\text { amplitude of motion } \\ \text { velocity } v & =\frac{d x}{d t} \end{aligned}$$
or $\frac{d x}{d t}=a(-\sin \omega t) \omega=-\omega a \sin \omega t$
or $v=-\omega a \sin \omega t$
$$=\omega \operatorname{acos}\left(\frac{\pi}{2}+\omega t\right) \quad\left[\because \sin \omega t=-\cos \left(\frac{\pi}{2}+\omega t\right)\right]$$
Now, phase of displacement $=\omega t$
Phase of velocity $=\frac{\pi}{2}+\omega t$
$\therefore$ Difference in phase of velocity to that of phase of displacement
$$=\frac{\pi}{2}+\omega t-\omega t=\frac{\pi}{2}$$