Answer :
Answer:
[tex]a=\dfrac{\pi}{3}[/tex]
Explanation:
given,
Amplitude ,A= 20 cm
distance between distance between two particle,y = 20 cm
Let equation of two SHM be x = A sin (ωt) and x = A sin (ωt + a)
where A is amplitude , ω is angular frequency and a is the phase difference.
now,
Distance between two particle at time t
y = A(sin(ωt+a)- sin (ωt))
using identity
[tex]sin A - sin B = 2sin(\dfrac{A-B}{2})cos(\dfrac{A+B}{2})[/tex]
[tex]y = A(2sin(\dfrac{a}{2})cos(\dfrac{2\omega t + a }{2}))[/tex]
for maxima [tex]cos(\dfrac{2\omega t+a}{2}) = 1[/tex]
maximum distance
[tex]y = A(2sin(\dfrac{a}{2}))[/tex]
[tex]20 = 20 \times 2 sin(\dfrac{a}{2})[/tex]
[tex]sin(\dfrac{a}{2})=\dfrac{1}{2}[/tex]
[tex]sin(\dfrac{a}{2})=sin(\dfrac{\pi}{6})[/tex]
[tex]\dfrac{a}{2}=\dfrac{\pi}{6}[/tex]
[tex]a=\dfrac{\pi}{3}[/tex]
hence, the phase difference between the two particle is equal to [tex]a=\dfrac{\pi}{3}[/tex]