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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.

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.

Grade:11

3 Answers

vikas askiitian expert
509 Points
10 years ago

 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

 

 

Simran more
15 Points
one year ago
Let amplitude of shm be A then by work energy theorem ,
         W=change in kinetic energy 
       mgx0-1/2 kx0^2=0
Hence mgx =kxo^2/2
Hence x=2mg/k
Thus, A=mg/k
Rishi Sharma
askIITians Faculty 646 Points
9 months ago
Dear Student,
Please find below the solution to your problem.

amplitude of block is A1 so from energy conservation
kA1^2 = Mv^2 ...........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)v1^2 /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
kx^2 /2 = (m+M)v1^2 /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

Thanks and Regards

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