# simple pendulum

• Sep 25th 2009, 06:27 AM
thereddevils
simple 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 AM
Hello thereddevils
Quote:

Originally Posted by thereddevils
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 .

The period of a simple pendulum is given by

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

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

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

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

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

• Sep 25th 2009, 07:19 AM
thereddevils
Quote:

Hello thereddevilsThe period of a simple pendulum is given by

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

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

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

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

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

Thanks Grandad . But i do not understand why is the accleration 9.81 +2

If the lift is accelerating upwards , but the accleration due to gravity is accelerating downwards ..
• Sep 25th 2009, 07:31 AM
The 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 $-2 \text{ ms}^{-2}$, and $g = +9.81$. So the difference is $9.81-(-2) = 11.81$.