JEE Main 2023 · 6 April · morning · Single correct · +4 / −1
A mass m is attached to two strings as shown in figure. The spring constants of two springs are K₁ and K₂. For the frictionless surface, the time period of oscillation of mass m is :
- (a) 2π √(m/(K₁ + K₂))
- (b) 2π √(m/(K₁ - K₂))
- (c) 1/2π√((K₁ + K₂)/m)
- (d) 1/2π√((K₁ - K₂)/m)
Answer and solution
Answer (a) 2π √(m/(K₁ + K₂))
Solution
Physics sir · 4 steps
Push the block one way: one spring pushes back, the other pulls back, so their stiffnesses add.
- Displace the block. Move m a small distance x to the right. The right spring K₁ is compressed by x and the left spring K₂ is stretched by x.
- Add the forces. The compressed K₁ pushes the block left with force K₁ x. The stretched K₂ pulls it left with force K₂ x. Both forces point back towards the mean position, so the net force is -(K₁ + K₂)x.
F = -(K₁ + K₂) x ⇒ K_(eff) = K₁ + K₂
- Time period. For any spring-mass system T = 2π√(m/K_(eff)). The surface is frictionless, so nothing else acts along the motion.
- So. T = 2π√(m/(K₁ + K₂)), which is option (a).
The idea to keep. A block between two springs fixed to opposite walls feels both restoring forces together, so the springs act in parallel and K_(eff) = K₁ + K₂, even though they look like they are in a line.
Answer from the NTA final answer key. Solution by PrepWiser.