Equation of SHM
Simple harmonic motion (SHM) is a specific type of oscillatory motion, in which
(a) particle moves in one dimension,
(b) particle moves to and fro about a fixed mean position (where 
(c) net force on the particle is always directed towards mean position, and
(d) magnitude of net force is always proportional to the displacement of particle from the mean position at that instant.
So, $$F_{n e t} \propto x \Rightarrow F_{n e t}=-k x$$
where, ‘k’ is known as force constant
$$
\begin{aligned}
& \Rightarrow \mathrm{ma}=-\mathrm{kx} \\
& \Rightarrow a=\frac{-k}{m} x \text { or } a=-\omega^2 x
\end{aligned}
$$
where, ω is known as angular frequency.
$$
\Rightarrow \frac{d^2 x}{d t^2}=-\omega^2 x
$$
This equation is called as the differential equation of S.H.M.
The general expression for x(t) satisfying the above equation is: $$\mathrm{x}(\mathrm{t})=\mathrm{A} \sin (\omega \mathrm{t}+\phi)$$
Velocity in SHM
At any instant t, $$, v(t)=A \omega \cos (\omega t+\phi)$$
At any position x, $$ v(x)=\pm \omega \sqrt{A^2-x^2}$$
Acceleration in SHM
At any instant t, $$a(t)=-\omega^2 A \sin (\omega t+\phi)$$
At any position x, $$a(x)=-\omega^2 x$$
Q.1. Prove that y = Aeiωt is an equation of SHM.
Solution:
According to given equation in problem, differentiating with respect to time we get,
$$
\frac{d y}{d t}=i A \omega e^{i \omega t}
$$
Differentiating again with respect to time, we get
$$
\frac{d^2 y}{d t^2}=-\omega^2 A e^{i \omega t}=-\omega^2 y \quad\left[\text { As } y=A e^{i \omega t}\right]
$$
Thus we have $$\frac{d^2 y}{d t^2}+\omega^2 y=0$$
This is the basic differential equation of SHM hence 
Q.2. Write the displacement equation representing the following conditions obtained in a simple harmonic motion. Amplitude =0.01 m, frequency = 600 Hz, initial phase = 
Solution:
Here $$\mathrm{A}=0.01 \mathrm{~m}, v=600 \mathrm{~Hz}, \quad \phi=\frac{\pi}{6}$$
The displacement equation of simple harmonic motion is
$$
\begin{aligned}
& \mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}+\phi)=\mathrm{A} \sin (2 \pi \nu \mathrm{t}+\phi) \\
& \Rightarrow \mathrm{y}=0.01 \sin \left(1200 \pi \mathrm{t}+\frac{\pi}{3}\right)
\end{aligned}
$$
Q.3. Find the amplitude and initial phase of a particle in SHM, whose motion equation is given as
$$
y=A \sin \omega t+B \cos \omega t
$$
Solution:
Here in the given equation we can write
$$
A=R \cos \phi
$$
and
$$
B=R \sin \phi
$$
Thus the given equation transforms to
$$
y=R \sin (\omega t+\phi)
$$
This is a general equation of SHM and here R is the amplitude of given SHM and 
Here ‘R’ is given by,
$$
R=\sqrt{A^2+B^2}
$$
Initial phase 
$$
\begin{aligned}
\tan \phi & =\frac{B}{A} \\
\phi & =\tan ^{-1}\left(\frac{B}{A}\right)
\end{aligned}
$$
Q.4. A body of mass 1 kg is executing simple harmonic motion which is given by $$y=6.0 \cos (100 t+4 \pi) \mathrm{cm}$$ What is the (i) amplitude of displacement, (ii) Angular frequency, (iii) initial phase, (iv) velocity, (v) acceleration, (vi) maximum kinetic energy?
Solution:
The given equation of SHM is
$$
y=6.0 \cos (100 t+\pi / 4) \mathrm{cm} .
$$
Comparing it with the standard equation ofSHM}, $$y=A sin(\omega t+\phi)$$, we have
(i) Amplitude $$A=6.0 \mathrm{~cm}$$
(ii) Angular frequency $$\omega=100 \mathrm{~s}^{-1}$$
(iii) Initial phase $$\phi=\pi / 4$$
(iv) Velocity $$v=\omega \sqrt{\left(A^2-y^2\right)}=100 \sqrt{\left(36-y^2\right)} \mathrm{cm} / \mathrm{s}$$
(v) Acceleration $$=-\omega^2 y=-(100)^2 y=-10^4 y$$
(vi) Kinetic energy $$=\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(A^2-y^2\right)$$
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