In the figure given below, a block of mass M=490 g placed on a frictionless table is connected with two springs having same spring constant (K=2 N m⁻¹). If the block is horizontally displaced through 'X' m then the number of complete oscillations it will make in 14 π seconds will be _____________.

Solution

Physics sir · 3 steps

Push the block: one spring stretches, the other compresses, both pull it back, so the stiffnesses add.

1. Effective spring constant. Displace the block by x: the left spring stretches by x and the right spring compresses by x. Both forces point back towards the mean position, so the restoring force is 2Kx. The springs act in parallel, giving k_(eff) = 2K = 4 N m⁻¹.

   T = 2π√(M/k_(eff)) = 2π√(0.49/4) = 2π × 0.7/2 = 0.7π s

2. Count oscillations. Use M = 490 g = 0.49 kg. The number of complete oscillations is the total time divided by one period: 14π/0.7π = 20.
3. So. The block makes 20 complete oscillations in 14π seconds.

The common slip: 10. Treating the springs as series gives k_(eff) = 1 N m⁻¹, a period of 1.4π s and 10 oscillations, not 20.

The idea to keep. A block between two springs fixed to opposite walls feels both springs act the same way, so they add like parallel springs: k_(eff) = k₁ + k₂. The amplitude never changes the period.

Answer from the NTA final answer key. Solution by PrepWiser.