Ответ:
pervonachalnaya dlina bila 20 sm
Объяснение:
l1 - pervonachalniy dlina
l2 - posledniy dlina
T1 - pervonachalnoy period
T2 - posledniy period.
mi znayem chto, l2=l1 + 60, T2 = 2T1,
![T1 = 2\pi \sqrt{ \frac{l1}{g} } T1 = 2\pi \sqrt{ \frac{l1}{g} }](https://tex.z-dn.net/?f=T1%20%3D%202%5Cpi%20%5Csqrt%7B%20%5Cfrac%7Bl1%7D%7Bg%7D%20%7D%20)
![T2 = 2\pi \sqrt{ \frac{l2}{g} } T2 = 2\pi \sqrt{ \frac{l2}{g} }](https://tex.z-dn.net/?f=T2%20%3D%202%5Cpi%20%5Csqrt%7B%20%5Cfrac%7Bl2%7D%7Bg%7D%20%7D%20)
otsuda mi naydyom
T2/T1=(l2/l1)^0.5, l2=4 l1
l1 + 60 = 4 l1
3 l1 = 60
l1 = 20 sm