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.

Result

l =  25 cm

Solution:

$x=2 \ \text{cm} \ \\ o=20 \ \text{cm} \ \\ o_{2}=o/2=20/2=10 \ \text{cm} \ \\ t^2=o_{2}^2 - x^2 \ \\ t=\sqrt{ o_{2}^2 - x^2 }=\sqrt{ 10^2 - 2^2 } \doteq 4 \ \sqrt{ 6 } \ \text{cm} \doteq 9.798 \ \text{cm} \ \\ l^2=(l-x)^2 + t^2 \ \\ l^2=l^2 -2lx + x^2 +t^2 \ \\ 2lx=x^2 +t^2 \ \\ l=(x^2 +t^2)/(2 \cdot \ x)=(2^2 +9.798^2)/(2 \cdot \ 2)=25 \ \text{cm}$

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Tiny

this helps me so much

2 years ago  Like

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