NTA Abhyas JEE Main2020PhysicsOscillationsPractice
The mass M shown in figure oscillates in simple harmonic motion with amplitude A . The amplitude of the point P is
Options
- Ak 1 A k 2
- Bk 2 A k 2 - k 1
- Ck 2 A k 1 + k 2
- Dk 2 A k 1
Correct answer
C. k 2 A k 1 + k 2
Step-by-step solution
If a force F is applied to M , say to the right, let A be the distance moved by M . If the system is released, it executes simple harmonic motion of amplitude A . If A 1 and A 2 ae the extensions in springs k 1 and k 2 then A = ( A 1   +   A 2 ) and F = k 1 A 1 = k 2 A 2 ⇒       A 1 = F k 1     and    A 2 = F k 2 ∴        A = A 1 + A 2 = F 1 k 1 + 1 k 2 = F k 1 + k 2 k 1 k 2 ⇒ &#