Bukhovtsev-et-al-Problems-in-Elementary-Physics

366 ANSWERS AND SOLUTIONS
623. Each half of the rod with a small sphere on its end is a mathematical
pendulum with a length d/2 that oscillates in the field of gravity of the
large sphere. In the fiel d of gravity of the Eart h the period of sma11 oscillations
of a mathematical pendulum is T o=2n V ~. According to the law
of gravitation mg=y m~E , and therefore
.. /fR2
To=2n V yM E
where 1'=6.67X 10- 8 cm 2/g·s2 is the gravity constant, ME is the mass of
the Earth, and R is the distance from the pendulum to the centre of the Earth.
Correspondingly, the period of small oscillations of the mathematical pendulum
with a length 1= ~ in the field of gravity of the large sphere will be
.. ;([[2
T=2n V 21'M::::: 5.4 hours
624. The period of oscillations of a mathematical pendulum is
T=2n V;,
where g' is the gravity acceleration in the corresponding coordinate system.
In our case
g'= JI g2+a 2
where g is the gravity acceleration with respect to the Earth.
Thus,
T=2n, /
V y g2+a 2
~_l__
625. T=2n .. I" 1 • Use the plus sign If the acceleration of the lilt is
V g±a
directed upward and the minus sign if it is directed downward.
626. The oscillations of the block in the cup are similar to those ot a
mathematical pendulum, with the only difference that instead of the tension
of the spring the block is acted upon by the reaction of the support. There..
fore, the sought oscillation period is
T=2n V:
F
627. When M ~ m, the acceleration of the cup is a=Ar-g. Therefore
(see Problem 626),