Pendulum

Calculate the length of the pendulum that is 2 cm lower in the lowest position than in the highest position. The length of the circular arc to be described when moving is 20cm.

Correct result:

l =  25 cm

Solution:

$$\begin{gathered} x=2\text{ cm } \\ o=20\text{ cm } \\ {o}_2=o\mathrm{/}2=20\mathrm{/}2=10\text{ cm } \\ {t}^2=o_2^2-x^2 \\ t=\sqrt{o_2^2-x^2}=\sqrt{10^2-2^2}=4\sqrt{6}\text{ cm}\doteq 9.798\text{ cm } \\ {l}^2=(l-x{)}^2+t^2 \\ {l}^2=l^2-2l_x+x^2+t^2 \\ 2l_x=x^2+t^2 \\ l=(x^2+t^2)\mathrm{/}(2\cdot x)=(2^2+9.798^2)\mathrm{/}(2\cdot 2)=25\text{ cm} \end{gathered}$$

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Tiny

this helps me so much

3 years ago  Like

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