The correct option is D Phase difference of two particles 18 cm apart at a given instant is π rad.
Given, equation of the wave is
y=8sinπ3(3t+x6)
Rewriting the above equation, we get
y(x,t)=8sin2π36(18t+x)
Comparing with the general equation of wave motion
y=Asin2πλ(vt+x)
Amplitude (A)=8 cm
Wavelength (λ)=36 cm
Wave velocity (v)=18 cm/s.
∴ frequency (f)=vλ=1836=0.5 Hz.
Now, phase of a particle at instant t1 is given by
ϕ1=π3(3t1+x6)
and at instant t2,
ϕ2=π3(3t2+x6)
The phase difference is given as
Δϕ=ϕ1−ϕ2=π3[3(t1−t2)]
For time difference Δt=0.4 s,
Δϕ=π3×3×0.4=180×0.4=72∘
Also, we know that Δϕ=2πλ|Δx|
Phase difference at an instant between two particles with path difference 18 cm is given by
Δϕ=2π36×18=π rad
[substituting the given data]
Hence, options (a) , (b) , (c) and (d) are the correct answers.