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A plank of mass M is placed on a smooth ...

A plank of mass `M` is placed on a smooth hroizonal surface. Two light identical springs each of stiffness `k` are rigidly connected to structs at the ends of the plank as shown. When the spring are in their unextended position the distance between their free ends is `3l`. a block of mass `m` is placed on the plank and pressed aganist one of the springs so that it is compressed by `l`. To keep the blocks at rest it is connected to the strut by means of a light string, initially the syetem is at rest. Now the string is burnt.

Time period of oscillation of block:

A

`(2pi +3) sqrt((2Mn)/(k(M+m)))`

B

`(pi+6)sqrt((2Mn)/(k(M+m)))`

C

`(pi+3)sqrt((Mm)/(k(M+m)))`

D

`(2pi+6)sqrt((Mm)/(k(M+m)))`

Text Solution

Verified by Experts

The correct Answer is:
D

(1) `CM` remains at rest, block moves `5t` on plank when system comes to rest
`-m [5l - Deltax] +M Deltax = 0, Deltax = (5ml)/(m+M)`
(2) In `CM` frame `(1)/(2) ((m.M)/(m+M)) v_(m//M)^(2) = (1)/(2) kl^(2)`
`V_(v//M) = sqrt(((M+m)/(Mm)))kl`
(3) Consider motion of block w.r.t. plank
`a_(M) = (kx)/(M)` right ward, `F =- (kx +ma_(M))` or
`(d^(2)x)/(dt^(2)) = ((k)/(m)x+a_(M)) = - k ((1)/(M)+(1)/(M))x omega^(2) = (pi)/(2) sqrt((Mm)/((m+m)k))`
Time to get the spring relaxed is `t_(1) = (pi)/(2) sqrt((Mm)/(M+m)k)`
Time taken by block to travel `3l` between springs
`=(3l)/(v_(m//M)) =(3l)/(sqrt((k(M+m))/(Mm))l)`
Time period of oscillation of block
`=2 [2t_(1)+ (3l)/(sqrt((k(M+m)Mm)/(Mm))l)] = (2pi +6) sqrt((Mm)/(k(M+m))`
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