1.A wave travelling in the positive xxx-direction with amplitude A=0.2A = 0.2A=0.2 m, velocity v=360v = 360v=360 m s−1^{-1}−1 and wavelength λ=60\lambda = 60λ=60 m, then the correct expression for the wave isa.y=0.2sin[2π(6t+x60)]y = 0.2\sin\left[2\pi\left(6t + \frac{x}{60}\right)\right]y=0.2sin[2π(6t+60x)]b.y=0.2sin[π(6t+x60)]y = 0.2\sin\left[\pi\left(6t + \frac{x}{60}\right)\right]y=0.2sin[π(6t+60x)]c.y=0.2sin[2π(6t−x60)]y = 0.2\sin\left[2\pi\left(6t - \frac{x}{60}\right)\right]y=0.2sin[2π(6t−60x)]d.y=0.2sin[π(6t−x60)]y = 0.2\sin\left[\pi\left(6t - \frac{x}{60}\right)\right]y=0.2sin[π(6t−60x)]Login to continueOnly logged in users canattempt or see the solution.