Question
A particle of mass 1   kg is subjected to a force which depends on the position as F → = - k x i ^ + y j ^   kg   m   s - 2 with k = 1   kg   s - 2 . At time t = 0 , the particle's position r → = 1 2 i ^ + 2 j ^   m and its velocity v → = - 2 i ^ + 2 j ^ + 2 π k ^   m   s - 1 . Let v x and v y denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When z = 0 . 5   m , the value of x v y - y v x is ______ m 2   s - 1 .
Step-by-step solution
Given here: F → = - k x i ^ + y j ^   kg   m   s - 2 and m = 1   kg In x-direction, F x = - x = m a x So, acceleration, a x = d 2 x d t 2 = - x Now, for particle executing SHM, displacement along x-direction is ⇒ x = A x sin ω t + ϕ x , here, angular frequency, ω = 1   rad   s - 1 and velocity, v x = A x ω cos ω t + ϕ x Given at t = 0 ,   x = 1 2   m and v x = - 2   m   s - 1 So, putting the values, we get 1 2 = A x sin ϕ x and - 2 = A x cos ϕ x From above two equations, ⇒ tan ϕ x = - 1 2           . . . 1 And A x = 5 2   m             . . . 2 Similarly, along y-direction, F y = - y = m a y , ⇒ a y = d 2 y d t 2 = - y So, displacement, y = A y sin ω t + ϕ y and velocity v y = A y ω cos ω t + ϕ y Given at t = 0 , y = 2   m and v y = 2   m   s - 1 So, putting the values, we get 2 = A y sin ϕ y and 2 = A y cos ϕ y From above two relations, we have ⇒ ϕ y = π 4         . . . 3 and A y = 2   m         . . . 4 Now, the value of x v y - y v x = 5 2 sin ω t + ϕ x × 2 cos ω t + ϕ y - 2 sin ω t + ϕ y × 5 2 cos ω t + ϕ x = 5 2 × 2 sin ω t + ϕ x cos ω t + ϕ y - sin ω t + ϕ y × cos ω t + ϕ x = 10 sin ϕ x - ϕ y = 10 sin ϕ x cos ϕ y - cos ϕ x sin ϕ y = 10 1 5 × 1 2 - - 2 5 × 1 2 = 3