IIT JEE , AIEEE Forum - Physics , Mechanics

rakhiagrawal

1)two simple pendulums of length 1m and 16m respectively r both given small displacement in the same direction at the same instant. they will be again in phase after the shorter pendulum has completed n oscilations .then n=
1)4/3
2)5/2
3)7
4)2

srini

HEY,

SINCE THE TIME PERIODS ARE IN THE RATIO TI/T2=1/4,

THE EQUATIONS CAN BE WRITTEN AS...

X=ASIN((2PI/T1)t)

X=ASIN((2PI/T2)t)

this will occur when x is same and TIME is same.

chetan_kp

is it 4/3
i will post soln if correct.....

rishipratimm

is the ans 4/3??

rakhiagrawal

@ chetan correct
1st option is the ans,........post the sol...

chetan_kp

let the time periods be T₁ and T₂

T = 2

( l/g)

T₁ / T_{2 =} 1/4

i.e. shorter pendulum moves 4 times faster than longer

now look at the diagram ...........

after the phase in the 3rd diagram the two pendulums come into the same phase after a short interval i.e. before the shorter pendulum completes the second oscillation

so no. of oscillations completed by the shorter should be between 1 and 2
in the given options it is clearly 4/3
this is just a shortcut i have found to save time(a lot of)
plz..... see if u can get me right or else nudge me

rakesh61

rakesh61

chetan_kp

rakhi here is the elaborated way
till 3rd diagram u can compute logically
after that consider this
y₁ = A Sin(

₁t)
y₂ = A Sin(

₂t +

)
y₁ = y₂
Sin (2

t/T) = Sin (2

t/4T +

/2) ......as second pendulum has initial phase

/2
solving u get t = T/3
in T sec 1st pendulum completes 1oscillation
so in T/3 it covers 1/3 oscillation
but in 1st , 2nd 3rd diagrams it has already covered 1 oscillation
so total oscillations = 1 + 1/3
= 4/3

nudge me for further doubts

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