JEE Main20206 Sep 2020Evening ShiftPhysicsOscillationsActual
When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion, is described by y t = y 0 sin 2 ωt , where ' y ' is measured from the lower end of upstretched spring. Then ω is :
Options
- A1 2 g y 0
- Bg y 0
- Cg 2 y 0
- D2 g y 0
Correct answer
C. g 2 y 0
Step-by-step solution
y = y 0 sin 2 ωt y = y 0 2 ( 1 - cot 2 ωt ) y - y 0 2 = - y 0 2   cos 2 ωt Y = A   cos 2 ωt 2 ω = k m maximum displacemnet = y 0 = mg k y 0 × ( 2 ω ) 2 = 2 g ω = g 2 y 0