Engineering Mechanics

Chapter 26 : Helical Springs and Pendulums „„„„„ 541

Let l = Length of the string,
t = Time period,
n = No. of beats in time t, and
r = Radius of the earth
We know that the value of g (i.e., acceleration due to gravity) varies inversely as r^2. Math-
ematically :

2

1

g

r

∝

∴ 22

1 k

gk

rr

=× = ...(where k is constant)

Taking logs of both sides,

log g = log k – 2 log r

Now taking differentials,

–2

dg dr

gr

=

or –
2

dg dr

gr

=

Now if h is height of the point where pendulum is placed above the earth’s surface, then dr = h.
Therefore

• 2

dg h

gr

=

We have studied in Art. 26.6. that

2

dg dn

gn

=

∴ –

dn h

nr

=

Notes : 1. Minus sign indicates that as dn
n

increases, the value of h

r

decreases and vice versa.

2. Similarly, it can be proved that if the value of h is taken as negative (i.e., the pendulum
   is taken below the surface of the earth as in the case of mines etc.) the value of dn/n will
   increase.
   Example 26.12. Find the approximate height of a mountain, at the top of which a pendulum,
   which beats seconds at sea level, loses 20 seconds a day. Take radius of the earth as 6400 km and
   the acceleration due to gravity varies as the square of the distance from the centre of the earth.

Solution. Given : No. of seconds the pendulum loses in one day (dn) = – 20 (Minus sign
due to loss of seconds) and radius of the earth (r) = 6400 km.

Let h = Height of mountain in km.

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

n = 24 × 60 × 60 = 86 400

We also know that –

dn h

nr

=