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 .
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
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