Question

2 particles move parallel to x axis about origin with same amplitude a and frequency f.At a certain instant they are found to be at distance a/3 from origin on opposite sides but their velocities in same direction.Phase difference between them

Answer 1

using standard equation of simple harmonic motion, $x=Asin\omega t$

case 1 : $x=\frac{a}{3}$, A = a

so, $\frac{a}{3}=asin\omega t$

$\implies sin\omega t=\frac{1}{3}$............(1)

case 2 : $x=-\frac{a}{3}$, A = a , and phase difference, $\phi$

so, $-\frac{a}{3}=asin(\omega t+\phi)$

or, $-\frac{1}{3}=sin(\omega t+\phi)$

or, $sin\omega t cos\phi+sin\phi cos\omega t$ = -1/3

from equation (1),

1/3 × cos$\phi$ + 2√2/3 × sin$\phi$ = -1/3

or, cos$\phi$+ 2√2sin$\phi$ = -1

after solving this equation we get,

cos$\phi$ = -1 , 7/9

$\phi$ = π or cos^-1(7/9)

but according to question, direction of velocities of both cases are same.

but this is not true when phase difference between them equals to π.

[ as you know , $v=\omega Acos\omega t$ , so, $v_1=\omega a cos\omega t$ , $v_2=\omega a cos(\omega t +\pi)$ = $-\omega a cos\omega t$ which shows opposite direction of first wave. hence, $\phi$ ≠ π ]

so, $\phi$ ≠ π

therefore, $\phi$ = cos^-1(7/9)