Question

A horizontal spring block system of mass 2kg executes SHM . When the block is passing through its equilibrium position, an object of mass 1kg is put on it and the 2 move together. The new amplitude of vibration is

Answer 1

Explanation:

Given A horizontal spring block system of mass 2kg executes SHM . When the block is passing through its equilibrium position, an object of mass 1kg is put on it and the 2 move together. The new amplitude of vibration is

• Since there is no energy loss 2v = 3v’
•                               Or v’ = 2v / 3
• So Ei = 1/2 m1v1^2
•         = 1/2 x 2 v^2
•         = v^2
• Ef = 1/2 m2v2^2
•      = 1/2 x 3 x v1^2
•       = 3/2 v1^2
• Since v’ = 2v / 3
•     = 3/2 (2v / 3)^2
•      = 2/3 v^2
• Now final energy is 1/2 KA’^2
• 1/2 K A’^2 = 2/3 V^2
•                 = 2/3 Ei
• Now Ef = 1/2 KA^2
• Equating both we get
• 1/2 KA’^2 = 2/3 x 1/2 KA^2
• Or A’^2 = 2/3 A^2
• Or A’ = √2/3 A

So the new amplitude of vibration is √2/3 A