Engineering Mechanics

Chapter 26 : Helical Springs and Pendulums „„„„„ 537

2. Law of mass. It states, “The time period (t) of a simple pendulum does not depend upon
   the mass of the body suspended at the free end of the string.”


3. Law of Length. It states, “The time period (t) of a simple pendulum is proportional to
   l, where l is the length of the string.”


4. Law of gravity. It states, “The time period (t) of a simple pendulum is inversely propor-
   tional to g. where g is the acceleration due to gravity.”
   Notes. The above laws of a simple pendulum are true from the equation of the time period i.e.

2·

l

t

g

=π

Example 26.8. A simple pendulum of amplitude 4° performs 24 oscillations in one minute.
Find (a) length of the pendulum (b) maximum acceleration of the bob, (c) maximum linear velocity
of the bob; and (d) maximum angular velocity of the bob.

Solution. Given : Angular amplitude (θ) = 4° =

4

180

π

rad Frequency (n) = 24 Hz and time

(t) = 1 min = 60 s.

(a) Length of the pendulum

Let l = Length of the pendulum

We know that time period for one oscillation,

60

2·5 s

24

t==

∴ 2·5 = 2 2

g9·8

ll

π=π

Squaring both sides of the equation

6·25 = (2 )^2

9·8

l

π

or 2

6·25 9·8

1·5 5 m

(2 )

l

×

==

π

Ans.

(b) Maximum linear acceleration of the bob

We know that maximum linear acceleration of the bob takes place, when it is at its extreme
position (or in other words, the displacement is maximum). We also know that displacement of the
bob (as per Fig. 26.6).

= Arc AC = OC × θ (in radians)

4

1.55 0·108 m

180

π

=× =

and angular velocity,

22

2·51 rads/s

t 2·5

ππ

ω= = =

∴ Maximum linear acceleration of the bob

amax = ω^2 × AC = (2·51)^2 × 0·108 = 0·68 m/s^2 Ans.

(c) Maximum linear velocity of the bob

We know that maximum linear velocity of the bob,

vmax = ω × AC = 2·51 × 0·108 = 0·27 m/s Ans.