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`        inside  a stationary  lift  the  time  period  of  a  simple  pendulum  is  4 sec  what  will  be the time period of the pendulum, if the  lift (1) moves upward with  an  acceleration  of  4m /sec ^2   (2)  moves downward  with an acceleration of  4m /sec ^2.`
2 years ago

Sanju
106 Points
```							Time period, T = 2π sqrt(l/g)Time period of the pendulum when the lift moves upward with an acceleration ‘a’ = 2π sqrt(l/(g+a))We have, T = 4sec = 2π sqrt(l/10)  ------------------(1)= 2π sqrt(l/(10+4))  -----------(2)’ , T2When acceleration upward with a= 4m/sDivide the equation (2) by (1) / 4] = [sqrt (10/14)]’[T  = 4* [sqrt (10/14)] = 3.38 sec2 [1] Therefore, time period of the pendulum when the lift moves upward with an acceleration of 4m/sTime period of the pendulum when the lift moves downward with an acceleration ‘a’ = 2π sqrt(l/(g-a))  = 4* [sqrt (10/6)] = 5.16 sec2 Proceeding as mentioned above, we get: [2] Time period of the pendulum when the lift moves upward with an acceleration of 4m/s
```
2 years ago
Sanju
106 Points
```							  = 4* [sqrt (10/14)] = 3.38 sec.................2 Time period of the pendulum when the lift moves upward with an acceleration of 4m/s  = 4* [sqrt (10/6)] = 5.16 sec.....2 Time period of the pendulum when the lift moves upward with an acceleration of 4m/s
```
2 years ago
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