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Banjit Das Grade: 11
        

A block with mass M attached to a horizontal spring with force contant k is moving with simple harmonic motion having amplitude A1.At the instant when the block passes through its equilibrium position a lump of putty with mass m is drropped vertically on the block from a very small height and sticks to it.


   (a) Find the new amplitude and period.


   (b) Repeat part (a) for the case in which the putty is dropped on the block when it at one end of its path.

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

 amplitude of block is A1 so from energy conservation


           kA12 = Mv2  ...........1


 


at equilibrium position velocity of block is v ...


 momentam of block is Mv..


after putty is dropped it sticks to it and now momentam of block + putty is (M+m)v1          (v1  is the new velocity of system..)


from conservation of momentam


                 Mv=(m+M)v1


                 v1 = Mv/(m+M) ...................2


 


Kinetic energy of this system is now , KE = (m+M)v12 /2     ...........3


                   due to this much kinetic energy of system spring gets compressed to distance x ....


 x is new amplitude of its motion....


          from energy conservation


        kx2 /2  =  (m+M)v12 /2


           x  = (M/m+M)1/2 A1                        


time period = T =2pi{(m+M)/k}1/2


 


for case b there is no change in amplitude but time period is T=2pi{(m+M)/k}1/2


 


 

6 years ago
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