(^540) „„„„„ A Textbook of Engineering Mechanics
We also know that
2
dn dg
ng

–1
86 400 2 301
dn

×
∴
86 400
––143·5s
2301
dn==
×
Ans.
Minus sign means that the pendulum will lose 143·5 seconds per day.
Example 26.11. Find the length of a pendulum, which will have one beat per second. If
such a pendulum loses 5 seconds a day, by how much length must it be shortened to keep the correct
time?
Solution. Given : Time (t) = 1 s and no. of second the pendulum loses in one day (dn) = – 5
(Minus sign due is loss of seconds.)
Length of the pendulum which will have one beat per second
Let l = Length of the pendulum.
We know that time for one beat (t).
1
9·8
ll
g
=π =π
Squaring both sides,
1 2
9·8
l
=π ×
∴ l==9·8 2 0·993m 993mm=
π
Ans.
Length, by which the pendulum should by shortened
Let dl = Length in mm by which the pendulum should be shortened,
to keep the correct time.
We know that no. of seconds in one day or 24 hours,
n = 24 × 60 × 60 = 86 400
We also know that –
2
dn dl
nl

–5

• 86 400 2 993

dl

=

×

∴

(^52993) 0·115 mm
86 400
dl==×× Ans.
It means that the length of the pendulum is 0·115 mm more than the correct length. Thus, for
correct time, the length of the pendulum should be shortened by 0·115 mm Ans.
26.7.GAIN OR LOSS IN THE NO. OF OSCILLATIONS DUE TO CHANGE IN
THE POSITION OF A SIMPLE PENDULUM
Consider a simple pendulum oscillating with a simple harmonic motion.