Question

a lift is ascending with acceleration g/3. what will be the time period of a simple pendulum suspended from the ceiling if its time period in stationary lift is T

Answer 1

luk in ascending lift g+a=g/3
a=2g/3
rit
then a=Aw^2sinwt
formula hai now ab time period k lliye mene equation bna li h par resting lift k eaquation smj ni a rhi

Answer 2

i want to give you a quick help. if elevator is ascending then always add g and acceleration of elevator.

if elevator is ascending then always subtract g and acceleration of elevator.

Answer 3

i know this tricks

Answer 4

bs ap mje resting lift k equation acceleration k terms m btao kaise likhu

Answer 5

simple heena g.

time period of the pendulum in rest condition, either in lift or else is T=2pisqrt(L/g)

Answer 6

anything else?

Answer 7

dusri equation to y bni meri
\[g \div3=A ^{2}4\times10\sin 2\pi t \div T\]

Answer 8

sory not g/3 its 2g/3

Answer 9

we have to compare if u luk the ques and we should always used the given quantities i think so

Answer 10

thnx JamesJ here. until i am looking someone else.

Answer 11

thnx alot heena.

Answer 12

If the lift is ascending at a rate of g/3, the mass of the pendulum experiences a net acceleration, a new apparent gravitational force, \( g' \), of
\[g' = g + g/3 = \frac{4}{3}g \]

You know
\[ T = 2\pi \sqrt{L/g} \]

Hence what you have to do is find the period \( T ' = 2\pi \sqrt{L/g'} \) in terms of \( T \).

Answer 13

ok thek hy madam.

Answer 14

lemme solve wait a sec

Answer 15

but here g'=2g/3 becoz apparent acc= a+g/3

Answer 16

Think about it. When a lift takes off upwards, i.e., accelerating upwards, you feel heavier, not lighter. Conversely when it is accelerating downwards you feel lighter.

Therefore we need to add g/3 to g, not subtract it.

Answer 17

i agree but not g we will add a
we were taught
in upwards a=g+a
in downwards a=g-a

Answer 18

no no u r rit i m telling the formula of tension
we will add g/3 to initial g
and thnx to u too shayaan

Answer 19

oh yes. you are welcome.
BTW how did you personaly mesg?

Answer 20

ary kesa bheja dear? mje notification milta hy. me is site pe new hn. mushkil se 1 month hua hy. mje itna kam ni ata.

Answer 21

okay if you want to through fundamentals then forget everything(formulas n tricks) and think of tension.
make a free body diagram of the pendulum
Fnet = T - mg
Fnet = ma = T - mg
m(g/3) = T-mg
T=4mg/3
since this tension decides the apparent weight of the bob, and hence the apparent
"g".. so apparent g= 4g/3
now,\[T= 2\pi \sqrt{l/g(apparent)}\]
therefore,\[T' = \sqrt{3}/2T\]