Manipal Entrance Test 2015 Physics Paper

We start a general form for a rightward moving wave,
y(x,t)=A‌sin(kx−ωt+ϕ) ...(i)
The given amplitude is A = 2 cm = 0.02 m
The wavelength is given as
λ=1m
Wave number =k=2π∕λ=2πm−1
Angular frequency,
ω=vk=10π‌rad∕s
From Eq. (i),
y(x,t)=(0.02)‌sin[2π(x−5t)+ϕ]
∵ For x = 0, t = 0
y=0 and

δy

δt

<0
i.e. 0.02‌sin‌ϕ=0 (as y = 0)
and −0.2π‌cos‌ϕ<0
From these conditions, we may conclude that
ϕ=2nπ, where n=0,2,4,6...
Therefore,
y(x,t)=(0.02m)‌sin[(2πm−1)x−(10πs−1)t]‌m