Engineering Mechanics

Chapter 26 : Helical Springs and Pendulums „„„„„ 539

t = Time period for one oscillation and

n = No. of beats or swings in time t.

We have seen in Art. 26.4 that the time period for one beat.

l

g

=π

∴Time-period for n beats tnl

g

=π

or

ttg

n

l l

g

==

π

π

...(i)

Taking logs of both sides,

1

log log – log (log – log )

2

nt=π+gl

Taking differential coefficients of variable (considering t and π as constants),

1

––

222

dn dg dl dg dl

nglgl

⎛⎞

==⎜⎟

⎝⎠

...(ii)

where dn is the change in time in n seconds.

Now, if the gravity changes (keeping length of pendulum constant), then

2

dn dg

ng

= ... 0

dl

l

⎛⎞=

⎜⎟

⎝⎠

Q

and if the length of pendulum changes (keeping g constant) then

• 2

dn dl

nl

= ...^0

dg

g

⎛⎞

⎜⎟=

⎝⎠

Q

Note. The minus sign indicates that as the value of dn/n increases, the value of dl/l decreases,
and vice versa. Or in other words, if dn increases, dl decreases (keeping n and l constant) i.e., if the
no. of beats in a clock increases, it is due to decrease in length of the pendulum. Similarly, if the no.
of beats decreases it means that it is due to increase in length of the pendulum.

Thus in order to set the pendulum correct, we have to reduce its length. Similarly, if the no.
of beats increase (or the pendulum gains time), it is due to decrease in the length of the pendulum.
Thus in order to set the pendulum correct, we have to increase its length.

Example 26.10. The gravity at the poles exceeds the gravity at the equator in the ratio of
301 : 300. If a pendulum regulated at the poles is taken to the equator, find how many seconds a day
will it lose?

Solution. Given : Gravity at pole (gp) = 301 and gravity at the equator (ge) = 300

∴ Change in gravity, when the pendulum is taken from the poles to the equator,

dg = 300 – 301 = – 1

Let dn = No. of seconds the pendulum will lose in one day.

We know that no. of seconds in one day or 24 hours

n = 24 × 60 × 60 = 86 400