# Analytical Treatment to SHM

Acceleration of Particle at any instant of time is

a = -ω2 x

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

$\large \int_{v_o}^{v} v dv = – \int_{x_o}^{x} \omega^2 x dx$

$\large v^2 – v_o^2 = -\omega^2 (x^2 – x_o^2)$

$\large v = \sqrt{(v_o^2 + \omega^2 x_o^2)-\omega^2 x^2 }$

$\large v = \omega \sqrt{(\frac{v_o^2}{\omega^2 } + x_o^2)-x^2}$

writing , $\large \frac{v_o^2}{\omega^2} + x_o^2 = A^2$

Equation becomes ,

$\large v = \omega \sqrt{A^2 -x^2}$

$\large \frac{dx}{dt} = \omega \sqrt{A^2 -x^2}$

$\large \int_{x_o}^{x}\frac{dx}{\sqrt{A^2 -x^2}} = \int_{0}^{t}\omega dt$

$\large [ sin^{-1}\frac{x}{A}]_{x_o}^{x} = \omega t$

$\large sin^{-1}\frac{x}{A} – sin^{-1}\frac{x_o}{A} = \omega t$

Writing , $\large sin^{-1}\frac{x_o}{A} = \phi$

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

$\large x = A sin(\omega t + \phi )$

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

$\large a = \frac{dv}{dt} = -\omega^2 A sin(\omega t + \phi )$

x as a function of time

x = A sin(ωt + φ)

For sin(ωt + φ) = 1 ⇒ x = A

For sin(ωt + φ) = -1 ⇒ x = -A

Therefore A is the maximum magnitude of displacement in the SHM and is known as amplitude of the oscillation .

Note that x can be positive or negative but A can never be negative.

#### Phase

The time varying quantity (ωt + φ) is referred as ‘ phase ‘ of the motion.

Denoting it with Φ, the equation of SHM becomes

x = A sin Φ . . . (10)

Importance of phase: By knowing phase, we can locate the position of particle as well as find the direction of subsequent motion.

Illustration : If two SHMs are represented by equations
y1 = 5[sin{2πt + (π/6)}] and y2 = 5[sin(3πt)+ √3cos(3πt)], find the ratio of their amplitudes.

Solution: We have, y2 = 5[sin(3πt)+ √3cos(3πt)]

Taking 5 = A cosθ . . . (1)

and 5√3 = Asinθ . . . (2)

We get,

y2 = Acosθsin3πt + Asinθcos3πt = Asin(3πt + π/3 )

From equation (1) and (2)

y2 = 10sin(3πt + π/3)

and so, A = 10 and θ = tan-1 (√3) = π/3

A1/A2 = 5/10 = 1/2

Exercise : In one oscillation of SHM, the particle can be at the same position for two different points of time. Like wise, can a particle have same phase for two different times in an oscillation?

Illustration : Find the phase of the motion in an SHM for x = A/2.

Solution : x = A sinφ

=> A/2 = A sinφ

=> φ = π/6 or 5π/6

Physical meaning of φ = π/6

Particle is at point P and it is going towards M.

Physical meaning of φ = 5π/6

Particle is at point P and it is going towards point O.

#### The constant φ

Φ = ωt + φ

at t = 0, Φ(0) = φ

Thus ‘ φ ‘ is the phase at the start of motion. (i.e. at the instant from where time is being counted).

φ is often referred as initial phase or phase constant or epoch.