Physics

Find the length of a pendulum that oscillates with a frequency of 0.19 Hz.

The acceleration due to gravity is 9.81 m/s 2 .

Formula:

For small amplitudes, the period of such a pendulum can be approximated by:

$\begin{array}{|lrcll|} \hline & T &=& 2\pi\sqrt{\frac{L}{g}} \\ \qquad L~ \text{expresses the pendulum length in meters } \\ \qquad g~ \text{expresses the acceleration of gravity}\approx 9.81 \frac{m}{s^2} \\ & T &=& \frac{1}{f} \\ \qquad f~ \text{expresses the frequency in Hz} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \frac{1}{f} &=& 2\pi\sqrt{\frac{L}{g}} \\ \frac{1}{2\pi f} &=& \sqrt{\frac{L}{g}} \\ \Big(\frac{1}{2\pi f}\Big)^2 &=& \frac{L}{g} \\ L &=& g\cdot \Big(\frac{1}{2\pi f}\Big)^2 \\ L &=& 9.81\cdot \left(\frac{1}{2\pi\cdot 0.19}\right)^2 \\ L &=& 9.81\cdot \left(0.83765759522\right)^2 \\ L &=& 9.81\cdot 0.70167024683 \\ \mathbf{L} &\mathbf{=}& \mathbf{6.88338512141\ m} \\ \hline \end{array}$

 The length of the pendulum is 6.88 m