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# 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...... 9 years ago

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.