## anonymous 4 years ago 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

1. anonymous

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

2. anonymous

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.

3. anonymous

i know this tricks

4. anonymous

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

5. anonymous

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

6. anonymous

anything else?

7. anonymous

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

8. anonymous

sory not g/3 its 2g/3

9. anonymous

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

10. anonymous

thnx JamesJ here. until i am looking someone else.

11. anonymous

thnx alot heena.

12. JamesJ

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 $$T$$.

13. anonymous

14. anonymous

lemme solve wait a sec

15. anonymous

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

16. JamesJ

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.

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

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.

21. anonymous

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|dw:1327858223624:dw| 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,$T' = \sqrt{3}/2T$