What is the frequency of the simple pendulum if its in the lift which is accelerating upwards at 2.00 m/s^2 . Given that the length of pendulum is 0.248 m .

I am not sure what 's the acceleration of gravity ..

Thanks .

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- Sep 25th 2009, 06:27 AMthereddevilssimple pendulum
What is the frequency of the simple pendulum if its in the lift which is accelerating upwards at 2.00 m/s^2 . Given that the length of pendulum is 0.248 m .

I am not sure what 's the acceleration of gravity ..

Thanks . - Sep 25th 2009, 07:03 AMGrandad
Hello thereddevilsThe period of a simple pendulum is given by

$\displaystyle T = 2\pi\sqrt{\frac{l}{g}}$

where $\displaystyle l$ is the length of the pendulum, and $\displaystyle g$ is the acceleration due to gravity.

If the lift is accelerating upwards at $\displaystyle 2\text{ ms}^{-2}$, then $\displaystyle g$ is effectively $\displaystyle 9.81+2 =11.81\text{ ms}^{-2}$. So with $\displaystyle l = 0.248$:

$\displaystyle T = 2\pi\sqrt{\frac{0.248}{11.81}}=0.9105$ sec

So the frequency $\displaystyle = \frac{1}{T}=1.098$ Hz

Grandad - Sep 25th 2009, 07:19 AMthereddevils
- Sep 25th 2009, 07:31 AMGrandad
Hello thereddevilsThe technical answer is that you need to find the difference between the vectors representing the two accelerations. So if we take the direction vertically downwards as positive, then the acceleration of the lift is $\displaystyle -2 \text{ ms}^{-2}$, and $\displaystyle g = +9.81$. So the difference is $\displaystyle 9.81-(-2) = 11.81$.

The more intuitive (and simpler) answer is to ask yourself: do you feel heavier or lighter when a lift is accelerating upwards? The answer, of course, is heavier. So the acceleration of the lift is*added*to the acceleration due to gravity.

Grandad