Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

Solutions 197

Therefore, using the formula for the period of a
simple pendulum, we can write

T 2n 1/

h

r (^) gsina '
where
Ii =
1112 1112
1/ 1; -I- 1! a 2 + b 2
sin a
a
V
412
a g •
1.100. The period of a simple pendulum is in-
versely proportional.to the square root of the free-
fall acceleration:
1
T o=.
g
Let the magnitude of the acceleration of the lift
be a. Then the period of the pendulum for the
lift moving upwards with an acceleration a will be
1
Tup oc
g F a
and for the lift moving downwards with the same
acceleration
Tdown cc

• 11 g — a

Obviously, the time measured by the pendulum

clock moving upwards with the acceleration a is

proportional to the ratio of the time tup of the

a 2 b 2 •

Consequently, the period of small oscillations of

the swing is

T= 2n