Mechanical Engineering Principles

SIMPLE HARMONIC MOTION 195

From equation (17.14),frequency,

f=

√

g

L

2 π

=

√

9. 81

2

2 π

= 0 .352 Hz

Problem 3. In order to determine the value

ofgat a certain point on the Earth’s surface,

a simple pendulum is used. If the pendulum

is of length 3 m and its frequency of

oscillation is 0.2875 Hz, determine the value

ofg.

From equation (17.14), frequency,

f=

√

g

L

2 π

i.e. 0. 2875 =

√

g

3

2 π

and ( 0. 2875 )^2 ×( 2 π)^2 =

g

3

3. 263 =

g

3

from which, acceleration due to gravity,

g= 3 × 3. 263 = 9 .789 m/s

17.5 The compound pendulum

Consider the compound pendulum of Figure 17.7,
which oscillates about the pointO. The pointGin
Figure 17.7 is the position of the pendulum’s centre
of gravity.
LetIo=mass moment of inertia aboutO

Now T=Ioα=−restoring couple

=−mghsinθ

From the parallel axis theorem,

IG=Io−mh^2 =mkG^2

or Io=mk^2 G+mh^2

whereIG=mass moment of inertia aboutG,

k^2 G=radius of gyration aboutG

O

h

x

y

ymg

h sin q

x

G

q

Figure 17.7

Hence

(

mk^2 G+mh^2

)

α=−mghsinθ

but for small displacements,

sinθ=θ

Hence, m

(

kG^2 +h^2

)

α=−mghθ

i.e.

(

k^2 G+h^2

)

α+ghθ= 0

or α+

gh

(k^2 G+h^2 )

θ= 0

However, α+ω^2 θ= 0

Therefore, ω^2 =

gh

(kG^2 +h^2 )

and ω=

√

gh

(kG^2 +h^2 )

( 17. 15 )

T=

2 π

ω

= 2 π

√

(k^2 G+h^2 )

gh

( 17. 16 )

and f=

1

T

=

1

2 π

√

gh

(k^2 G+h^2 )

( 17. 17 )

Problem 4. It is required to determine the

mass moment of inertia aboutGof a metal

ring, which has a complex cross-sectional

area. To achieve this, the metal ring is

oscillated about a knife edge, as shown in