2019-03-01_Physics_Times

3 7

2 5

2 2

t

 

   

7

8

T

 t

For n 7

3

7

2

t t



    

11 11

2

2 8

T

t t



   

So the two particles meet at

3 7 11

, ,

8 8 8

T T T

etc

Finding the times at 2nd, 3rd meeting is very

difficult with the first two methods.

Two pendulums whose lengths are 100cm and

121cm are suspended side by side.Their bobs are

pulled together and then released. After how many

minimum oscillations of the longer pendulum, will

the two be in phase again.

Let T 1 andT 2 are the time periods of two

Pendulums 1

100

T 2

g

  and 2

121

T 2

g

 

(T T 2  1 becausel l 2  1 ).

Let at t 0 they start swinging together. Since

their time periods are different, the swinging will

not be in unison always. Only when number of

completed oscillation differ by an integer, the two

pendulums will again begin to swing together.

Let longer length pendulum complete N

oscillations and shorter length pendulum complete

(N+1) oscillations, for the unison swinging, then

( 1)N T NT  1 2

100 121

( 1) 2N N 2 N 10

g g

     

Let  0 is the angular amplitude of the two

pendulums.   1 and 2 are angular frequencies of

the two pendulums

1 100 & (^2121)
g g
  
(T T 1   2 &  1 2 )
The angular displacements of the two pendulums
as a function of time are.
      1  0 cos 1 t& 2  0 cos( ) 2 t
Since both start from extreme position.
When the two pendulums meet again then
    1    2 cos 1 t tcos 2
   1 t n t  (^22)
For n 1
1 2
2
t

 


Let N be the number of oscillations of longer
pendulum so that t NT 2
2
1 2 2 1 2
2 2
N N
  
    
  
 
2
1 2
/
/ /
g l
N
g l g l


1
2 1
l
l l
100
121 100


10
10,
21
N
The corresponding time for N 10 is
1 2
2
t 22.45s

 
 

The corresponding time for
10
21
N is
1 2
2
t 1.

 
 

It is interesting to note that the two pendulums
meet each other several times in the time interval of
22.45 s
For N=1, 2, 3 - - -
Note:
3.
Sol:
Method-1:
Method-2^121212
2 4 6
t , ,
  
     

   etc 
The corresponding times are
t s s s1.046 , 2.084 , 3.127 etc
At these times the two pendulums will have
different phases.