Mechanical Engineering Principles

192 MECHANICAL ENGINEERING PRINCIPLES

Equation (17.6) shows that the acceleration along
the lineyyis directly proportional to the displace-
ment along this line, therefore the pointCis moving
with SHM. Now

T=

2 π

ω

,

but from equation (17.6),aC=ω^2 y

i.e. ω^2 =

a

y

Therefore,

T=

2 π

√

a

y

or T= 2 π

√

y

a

i.e. T= 2 π

√

dispacement

acceleration

In general, from equation (17.6),

a+ω^2 y= 0 ( 17. 7 )

17.3 The spring-mass system

(a) Vibrating horizontally

Consider a massmresting on a smooth surface
and attached to a spring of stiffnessk, as shown
in Figure 17.2.

x

kx

Figure 17.2

If the mass is given a small displacementx,the
spring will exert a resisting force ofkx,

i.e. F=−kx

But, F=ma,

hence, ma=−kx

or ma+kx= 0

or a+

k

m

x= 0 ( 17. 8 )

Equation (17.8) shows that this mass is oscil-

lating (or vibrating) in SHM, or according to

equation (17.7). Comparing equation (17.7) with

equation (17.8) we see that

ω^2 =

k

m

from which, ω=

√

k

m

Now T=

2 π

ω

= 2 π

√

m

k

andf=frequency of oscillation or vibration

i.e. f=

ω

2 π

=

1

2 π

√

k

m

( 17. 9 )

(b) Vibrating vertically

Consider a massm, supported by a vertical spring

of stiffness k, as shown in Figure 17.3. In this

equilibrium position, the mass has an initial down-

ward static deflection ofyo. If the mass is given

an additional downward displacement ofyand then

released, it will vibrate vertically.

y 0

y

Figure 17.3

The force exerted by the spring=−k(yo+y)

Therefore, F=mg−k(yo+y)=ma

i.e. F=mg−kyo−ky=ma