JEE Advanced2019PhysicsLaws of MotionActual
A block of mass 2 M is attached to a massless spring with spring-constant k . This block is connected to two other blocks of masses M and 2 M using two massless pulleys and strings. The accelerations of the blocks are a 1 , a 2 and a 3 as shown in figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is x 0 . Which of the following option(s) is/are corr
Options
- Ax 0 = 4 M g k
- BWhen spring achieves an extension of x 0 2 for the first time, the speed of the block connected to the spring
- Ca 2 - a 1 = a 1 - a 3
- DAt an extension of x 0 4 of the spring, the magnitude of acceleration of the block connected to the spring is
Correct answer
C. a 2 - a 1 = a 1 - a 3
Step-by-step solution
Using string constraint a 1 = a 2 + a 3 2 ⇒ 2 a 1 = a 2 + a 3 a 1 - a 3 = a 2 - a 1 ⇒ C is correct. for other options use m equivalent Equivalent mass Or Reduced mass m e q = 2 m 1 × m 2 m 1 + m 2 (for a two-mass system) = 2 2 m × m 2 m + m = 4 m 3 ⇒ T = 4 m 3 . g ∴ 2 T = 2 . 4 m g 3 = 8 m 3 g ⇒ m e q = 8 m 3 Using mechanical energy conservation ⇒ 1 2 k x 0 2 = 8 m g 3 x 0 x 0 = 16 m g 3 k ⇒ A is incorrect Option B ⇒ V x 0 2 = V m a x = x 0 2 ω = x 0 2 k 2 m + 8 m 3 = x 0 2 3 k 14 m = g 32 21 k Option D ⇒ a x 0 4 =