Two identical springs of spring constant $k$ are attached to a block of mass $m$ and to fixed supports as shown in figure. When the mass is displaced from equilibrium position by a distance $x$ towards right, find the restoring force.
Consider the diagram in which the block is displaced right through x.
The right spring gets compressed by $x$ developing a restoring force $k x$ towards left on the block. The left spring is stretched by an amount $x$ developing a restoring force $k x$ towards left on the block as given in the free body diagram of the block.
$$\begin{aligned} \text { Hence, total force (restoring) } & =(k x+k x) \quad[\because \text { Both forces are in same direction }] \\ & =2 k x \text { towards left } \end{aligned}$$
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 $=\frac{4 A}{A}=4$