can you tell me the problem number 30 in shm chapter in hc verma part 1 ....even i see the solution i am not able tp understand please explain......

SOLUTION

Let the amplitude of oscillation of m and M be x₁ and x₂ respectievely

a) From the law of conservation of momentum :-

mx₁ = Mx₂ ...........(1) [bcoz internal forces are present]

Again , 1/2kx₀² = 1/2k (x₁ + x₂)²

Therefore x₀ = x₁ + x₂

[Block and mass oscillates in opposite direction. But x----> stretched part]

From equations (1) and (2) :-

x₀ =x₁ +(m/M)x₁ = [(m+M)/M]x₁

Therefore x₁(Mx₀/m+M)

So x₂ = x₀ -x₁

         = mx₀/m+M

b) At any positions , let v₁ and v₂ be the velocities

where v₁ = velocity of m w.r.t M

By energy method :-

Total Energy = Constant

(1/2)Mv² + 1/2m(v₁ - v₂)² +1/2k(x₁ - x₂)² = constant ........(1)

By using the law of conservation of momentum we get :-

x₁ = (M/m)x₂ .............(2)

v₁ - v₂ = (M/m)v₂...................(3)

Putting the values in (1) we get:-

mv₂² + k(1 + M/m)x₂² = constant

Taking derivative from both sides we get :-

w=√[k(m+M)/Mm]

Therefore Time period will be 2∏w

On putting the value of w you will get the answer.