JEE Main 2007Annual, Paper 1PhysicsMediumOscillations and Waves185× since 2002A point mass oscillates along the xxx-axis according to the law x=x0 cos(ωt−π/4).x = {x_0}\,\cos \left( {\omega t - \pi /4} \right).x=x0cos(ωt−π/4). If the acceleration of the particle is written as a=A cos(ωt+δ),a = A\,\cos \left( {\omega t + \delta } \right),a=Acos(ωt+δ), thenAOption A. A=x0ω2, δ=3π/4A = {x_0}{\omega ^2},\,\,\delta = 3\pi /4A=x0ω2,δ=3π/4BOption B. A=x0, δ=−π/4A = {x_0},\,\,\delta = - \pi /4A=x0,δ=−π/4COption C. A=x0ω2, δ=π/4A = {x_0}{\omega ^2},\,\,\delta = \pi /4A=x0ω2,δ=π/4DOption D. A=x0ω2, δ=−π/4A = {x_0}{\omega ^2},\,\,\delta = - \pi /4A=x0ω2,δ=−π/4CheckSkip · show solution
