Engineering Mechanics

Chapter 26 : Helical Springs and Pendulums „„„„„ 529

Let a body be attached to the lower end. Let A-A be the equilibrium
position of the spring, after the body is attached. If the spring is stretched
up to B-B and then released, the body will move up and down with a
simple harmonic motion.

Let m = Mass of the body in kg (such that

its weight (W) is mg newtons)

s = Stiffness of the spring in N/m

x = Displacement of the load below

the equilibrium position in metres.

a = Acceleration of the body in m/s^2

g = Gravitational acceleration, and

t = Periodic time.

We know that the deflection of spring,

Wmg

ss

δ= =

Then disturbing force = Mass × Acceleration = m.a ...(i)

and restoring force = sx ...(ii)

Equating equations (i) and (ii),

ma = sx or mx

sa

= ...(iii)

We know that in simple harmonic motion, time period,

Displacement

22

Acceleration

x

t

a

=π =π 2

m

s

=π

2

2

g

πδ

=π

ω

2

...( and )

mg

t

s

π

==δ

ω

Q

∴

g

ω=

δ

where ω is the angular velocity in rad/ sec.

Notes. 1. Frequency of motion,

11 1

22

sg

n

tm

== =

ππδ

2. From equation (iii) we find that
   ma
   x
   s

=

Thus we see that x is directly proportional to m/s.
Example 26.1. A 4 kg mass hung at one end of a helical spring and is set vibrating
vertically. The mass makes 2 vibrations per second. Determine the stiffness of the spring.

Solution. Given : Mass (m) = 4 kg and frequency (n) = 2 vib/s = 2 Hz

Let s = Stiffness of the spring.

Fig. 26.1 Helical spring.