NEET2026PhysicsChapterActual
A block of mass 4 kg is executing simple harmonic motion on a smooth horizontal surface with an amplitude A and frequency , under the restoring force of an ideal spring. When the block passes through its mean position, a lump of putty of mass 5 kg is dropped vertically onto the block and sticks to it. What will be the new frequency ( ' ) and new amplitude ( A' ) of the oscillation?
Options
- A' = 3 2 , A' = 2 3 A
- B' = 2 3 , A' = 4 9 A
- C' = 2 3 , A' = 2 3 A
- D' = 3 2 , A' = 4 9 A
Correct answer
C. ' = 2 3 , A' = 2 3 A
Step-by-step solution
The initial frequency of the block is given by: = 1 2 k m When the putty of mass 5 kg is dropped on the block of mass 4 kg , the new total mass becomes M = 4 + 5 = 9 kg . The new frequency ' is: ' = 1 2 k M = 1 2 k 9 Taking the ratio of the new frequency to the old frequency: ' = m M = 4 9 = 2 3 ' = 2 3 At the mean position, the velocity of the block is maximum. Let the initial velocity be v . v = A = A(2 ) Since the putty is dropped vertically, there is no external force in the horizontal direction. By conservatio