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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

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  1. anonymous
    • 4 years ago
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    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. Shayaan_Mustafa
    • 4 years ago
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    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
    • 4 years ago
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    i know this tricks

  4. anonymous
    • 4 years ago
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    bs ap mje resting lift k equation acceleration k terms m btao kaise likhu

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

  6. Shayaan_Mustafa
    • 4 years ago
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    anything else?

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

  8. anonymous
    • 4 years ago
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    sory not g/3 its 2g/3

  9. anonymous
    • 4 years ago
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    we have to compare if u luk the ques and we should always used the given quantities i think so

  10. Shayaan_Mustafa
    • 4 years ago
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    thnx JamesJ here. until i am looking someone else.

  11. Shayaan_Mustafa
    • 4 years ago
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    thnx alot heena.

  12. JamesJ
    • 4 years ago
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    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. Shayaan_Mustafa
    • 4 years ago
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    ok thek hy madam.

  14. anonymous
    • 4 years ago
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    lemme solve wait a sec

  15. anonymous
    • 4 years ago
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    but here g'=2g/3 becoz apparent acc= a+g/3

  16. JamesJ
    • 4 years ago
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    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
    • 4 years ago
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    i agree but not g we will add a we were taught in upwards a=g+a in downwards a=g-a

  18. anonymous
    • 4 years ago
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    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. Shayaan_Mustafa
    • 4 years ago
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    oh yes. you are welcome. BTW how did you personaly mesg?

  20. Shayaan_Mustafa
    • 4 years ago
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    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
    • 4 years ago
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    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,\[T' = \sqrt{3}/2T\]

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