JEE Main202122 Jul 2021Morning ShiftPhysicsLaws of MotionActual
The motion of a mass on a spring, with spring constant K is as shown in figure. The equation of motion is given by, x ( t ) = A sin ω t + B cos ω t with ω = K m . Suppose that at time t = 0 , the position of mass is x ( 0 ) and velocity v ( 0 ) , then its displacement can also be represented as x ( t ) = C cos ( ω t - ϕ ) , where C and ϕ are
Options
- AC = 2 v ( 0 ) 2 ω 2 + x ( 0 ) 2 ,   ϕ = tan - 1 v ( 0 ) x ( 0 ) ω
- BC = 2 v ( 0 ) 2 ω 2 + x ( 0 ) 2 ,   ϕ = tan - 1 x ( 0 ) ω 2 v ( 0 )
- CC = v ( 0 ) 2 ω 2 + x ( 0 ) 2 ,   ϕ = tan - 1 x ( 0 ) ω v ( 0 )
- DC = v ( 0 ) 2 ω 2 + x ( 0 ) 2 ,   ϕ = tan - 1 v ( 0 ) x ( 0 ) ω
Correct answer
D. C = v ( 0 ) 2 ω 2 + x ( 0 ) 2 ,   ϕ = tan - 1 v ( 0 ) x ( 0 ) ω
Step-by-step solution
x = A s i n ω t + B c o s ω t v = d x d t = A ω cos ω t - B ω sin ω t At t = 0 ,   x ( 0 ) = B v ( 0 ) = A ω x = A sin ω + B sin ω t + 90 ° A net = A 2 + B 2 tan α = B A ⇒ cot α = A B ⇒ x = A 2 + B 2 sin ( ω t + α ) ⇒ x = A 2 + B 2 cos ( ω t - ( 90 - α ) ) x = C cos ( ω t - ϕ ) ⇒ C = A 2 + B 2 C = [ v ( 0 ) ] 2 ω 2 + [ x ( 0 ) ] 2 ϕ = 90 - α tan α = cos α = A B ⇒ tan