A community for students.
Here's the question you clicked on:
 0 viewing
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
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

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0luk 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0bs ap mje resting lift k equation acceleration k terms m btao kaise likhu

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0simple heena g. time period of the pendulum in rest condition, either in lift or else is T=2pisqrt(L/g)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dusri equation to y bni meri \[g \div3=A ^{2}4\times10\sin 2\pi t \div T\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sory not g/3 its 2g/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we have to compare if u luk the ques and we should always used the given quantities i think so

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thnx JamesJ here. until i am looking someone else.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3If 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 \( T \).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lemme solve wait a sec

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but here g'=2g/3 becoz apparent acc= a+g/3

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.3Think 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i agree but not g we will add a we were taught in upwards a=g+a in downwards a=ga

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yes. you are welcome. BTW how did you personaly mesg?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ary kesa bheja dear? mje notification milta hy. me is site pe new hn. mushkil se 1 month hua hy. mje itna kam ni ata.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay if you want to through fundamentals then forget everything(formulas n tricks) and think of tension. make a free body diagram of the pendulumdw:1327858223624:dw Fnet = T  mg Fnet = ma = T  mg m(g/3) = Tmg 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,\[T' = \sqrt{3}/2T\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.