Question

derive the equation for a simple harmonic wave in one dimension and deduce a relation for the phase difference between two points of on a wave?​

Answer 1

for a simple harmonic motion, acceleration of particle should be act opposite direction of its motion and also acceleration should be directly proportional to displacement from its mean position .

e.g., $a=-\omega^2x$

we can write, a = v dv/dx

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

or, $\int\limits^0_v{v}\,dv=-\omega^2\int\limits^A_x{x}\,dx$

maximum displacement from mean position is known as amplitude so, Here upper limit of position will be A.

and at extreme position velocity of particle will be zero. so, upper limit of velocity will be zero.

or, $v^2=\omega^2(A^2-x^2)$

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

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

or, $\int{\frac{dx}{\sqrt{A^2-x^2}}}=\omega\int\limits^t_{t_0}{dt}$

or, $sin^{-1}\frac{x}{A}=\omega t+\omega t_0$

here, $\omega t_0$ is phase difference.e.g., $\phi$

or, $x=Asin(\omega t+\phi)$ this is the standard equation of simple harmonic motion.

relation between phase difference and path difference is ;

$\phi=\frac{2\pi}{\lambda}\Delta x$