What are the two basic characteristics of a simple harmonic motion?
The two basic characteristics of a simple harmonic motion
(i) Acceleration is directly proportional to displacement.
(ii) The direction of acceleration is always towards the mean position, that is opposite to displacement.
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.