BITSAT2015PhysicsLaws of MotionActual
A load of mass m falls from a height h on to the scale pan hung from the spring as shown in the figure. If the spring constant is k and mass of the scale pan is zero and the mass m does not bounce relative to the pan, then the amplitude of vibration is
Options
- Amg / d
- Bmg k ( 1+2 hk mg )
- Cmg k + mg k ( 1+2 hk mg )
- Dmg k ( 1+2 hk mg - mg k )
Correct answer
B. mg k ( 1+2 hk mg )
Step-by-step solution
According to energy conservation principle, If, x ₁ is maximum elongation in the spring when the particle is in its lowest extreme position. Then, array l mgh = 1 2 kx ₁²- mgx ₁ 1 2 kx ₁²- mgx ₁- mgh =0 or, x ₁²- 2 mg k x ₁- 2 mg k h =0 x ₁= 2 mg k . ( 2 mg k )²+4 2 mg k h ] 2 array Amplitude A = X ₁- X ₀ (elongation in spring for equilibrium position) A = mg k (1+ 2 hk mg )