1.2 Shm equation  (Page 2/2)

Angular acceleration, $\alpha =-\frac{mgl}{I}\theta$ ------ for rotational SHM of pendulum

where “m” and “I” are mass and moment of inertia of the oscillating objects in two systems.

In order to understand the nature of cause, we focus on the block-spring system. When the block is to the right of origin, “x” is positive and restoring spring force is negative. This means that restoring force (resulting from elongation of the spring) is directed left towards the neutral position (center of oscillation). This force accelerates the block towards the center. As a result, the block picks up velocity till it reaches the center.

The plot, here, depicts nature of force about the center of oscillation bounded between maximum displacements on either side.

As the block moves past the center, “x” is negative and force is positive. This means that restoring force (resulting from compression of the spring) is directed right towards the center. The acceleration is positive, but opposite to direction of velocity. As such restoring force decelerates the block.

In the nutshell, after the block is released at one extreme, it moves, first, with acceleration up to the center and then moves beyond center towards left with deceleration till velocity becomes zero at the opposite extreme. It is clear that block has maximum velocity at the center and least at the extreme positions (zero).

From the discussion, the characterizing aspects of the restoring force responsible for SHM are :

• The restoring force is always directed towards the center of oscillation.
• The restoring force changes direction across the center.
• The restoring force first accelerates the object till it reaches the center and then decelerates the object till it reaches the other extreme.
• The process of “acceleration” and “deceleration” keeps alternating in each half of the motion.

Differential form of shm equation

We observed that acceleration of the object in SHM is proportional to negative of displacement. Here, we shall formulate the general equation for SHM in linear motion with the understanding that same can be extended to SHM along curved path. In that case, we only need to replace linear quantities with corresponding angular quantities.

$$a=-{\omega }^{2}x$$

where “ ${\omega }^2$ ” is a constant. The constant “ω” turns out to be angular frequency of SHM. This equation is the basic equation for SHM. For block-spring system, it can be seen that :

$$\omega =\sqrt{\left(\frac{k}{m}\right)}$$

where “k” is the spring constant and “m” is the mass of the oscillating block. We can, now, write acceleration as differential,

$$\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$$

$$\Rightarrow \frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$$

This is the SHM equation in differential form for linear oscillation. A corresponding equation of motion in the context of angular SHM is :

$$\Rightarrow \frac{d^2\theta }{d{t}^2}+{\omega }^2\theta =0$$

where "θ" is the angular displacement.

Solution of shm differential equation

In order to solve the differential equation, we consider position of the oscillating object at an initial displacement $x_0$ at t =0. We need to emphasize that “ $x_0$ ” is initial position – not the extreme position, which is equal to amplitude “A”. Let

$$t=\mathrm{0,}x=x_{\mathrm{0,}}v=v_0$$

We shall solve this equation in two parts. We shall first solve equation of motion for the velocity as acceleration can be written as differentiation of velocity w.r.t to time. Once, we have the expression for velocity, we can solve velocity equation to obtain a relation for displacement as its derivative w.r.t time is equal to velocity.

Velocity

We write SHM equation as differential of velocity :

$$a=\frac{dv}{dt}==-{\omega }^{2}x$$

$$\Rightarrow \frac{dv}{dx}X\frac{dx}{dt}=-{\omega }^{2}x$$

$$\Rightarrow v\frac{dv}{dx}=-{\omega }^{2}x$$

Arranging terms with same variable on either side, we have :

$$\Rightarrow vdv=-{\omega }^{2}xdx$$

Integrating on either side between interval, while keeping constant out of the integral sign :

$$\Rightarrow \underset{v_0}{\overset{v}{\int }}vdv=-{\omega }^2\underset{x_0}{\overset{x}{\int }}xdx$$

$$\Rightarrow [\frac{v^2}{2}\underset{v_0}{\overset{v}{]}}=-{\omega }^2[\frac{x^2}{2}\underset{x_0}{\overset{x}{]}}$$

$$\Rightarrow \left(v^2-v_0^2\right)=-{\omega }^2\left(x^2-x_0^2\right)$$

$$\Rightarrow v=\sqrt{\{\left(v_0^2+{\omega }^2{x}_0^2\right)-{\omega }^2{x}^2\}}$$

$$\Rightarrow v=\omega \sqrt{\{\left(\frac{v_0^2}{{\omega }^2}+x_0^2\right)-x^2\}}$$

We put $\left(\frac{v_0^2}{{\omega }^2}+x_0^2\right)=A^2$ . We shall see that “A” turns out to be the amplitude of SHM.

$$\Rightarrow v=\omega \sqrt{\left(A^2-x^2\right)}$$

This is the equation of velocity. When x = A or -A,

$$\Rightarrow v=\omega \sqrt{\left(A^2-A^2\right)}=0$$

when x = 0,

$$\Rightarrow v\max =\omega \sqrt{\left(A^2-0^2\right)}=\omega A$$

Displacement

We write velocity as differential of displacement :

$$v=\frac{dx}{dt}=\omega \sqrt{\left(A^2-x^2\right)}$$

Arranging terms with same variable on either side, we have :

$$\Rightarrow \frac{dx}{\sqrt{\left(A^2-x^2\right)}}=\omega dt$$

Integrating on either side between interval, while keeping constant out of the integral sign :

$$\Rightarrow \underset{x_0}{\overset{x}{\int }}\frac{dx}{\sqrt{\left(A^2-x^2\right)}}=\omega \underset{0}{\overset{t}{\int }}dt$$

$$\Rightarrow [{\sin }^{-1}\frac{x}{A}\underset{x_0}{\overset{x}{]}}=\omega t$$

$$\Rightarrow {\sin }^{-1}\frac{x}{A}-{\sin }^{-1}\frac{x_0}{A}=\omega t$$

Let ${\sin }^{-1}\frac{x_0}{A}=\phi$ . We shall see that “φ” turns out to be the phase constant of SHM.

$$\Rightarrow {\sin }^{-1}\frac{x}{A}=\omega t+\phi$$

$$\Rightarrow x=A\sin \left(\omega t+\phi \right)$$

This is one of solutions of the differential equation. We can check this by differentiating this equation twice with respect to time to yield equation of motion :

$$\frac{dx}{dt}=A\omega \cos \left(\omega t+\phi \right)$$

$$\frac{d^{2}x}{d{t}^2}=-A{\omega }^2\sin \left(\omega t+\phi \right)=-{\omega }^{2}x$$

$$\Rightarrow \frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$$

Similarly, it is found that equation of displacement in cosine form, $x=A\cos \left(\omega t+\phi \right)$ , also satisfies the equation of motion. As such, we can use either of two forms to represent displacement in SHM. Further, we can write general solution of the equation as :

$$x=A\sin \omega t+B\cos \omega t$$

This equation can be reduced to single sine or cosine function with appropriate substitution.

Example

Problem 1: Find the time taken by a particle executing SHM in going from mean position to half the amplitude. The time period of oscillation is 2 s.

Solution : Employing expression for displacement, we have :

$$x=A\sin \omega t$$

We have deliberately used sine function to represent displacement as we are required to determine time for displacement from mean position to a certain point. We could ofcourse stick with cosine function, but then we would need to add a phase constant “π/2” or “-π/2”. The two approach yields the same expression of displacement as above.

Now, according to question,

$$\Rightarrow \frac{A}{2}=A\sin \omega t$$

$$\Rightarrow \sin \omega t=\frac{1}{2}=\sin \frac{\pi }{6}$$

$$\Rightarrow \omega t=\frac{\pi }{6}$$

$$\Rightarrow t=\frac{\pi }{6\omega }=\frac{\pi T}{6X2\pi }=\frac{T}{12}=\frac{2}{12}=\frac{1}{6}s$$