Mechanical Engineering Principles

SIMPLE HARMONIC MOTION 193

But,kyo=mg, hence F=mg−mg−ky=ma

Thus, ma+ky= 0

or a+

k

m

y= 0

i.e. SHM takes place and periodic time,

T= 2 π

√

m

k

( 17. 10 )

and frequency, f=

ω

2 π

=

1

2 π

√

k

m

( 17. 11 )

as before (from equation (17.9)).
Comparing equations (17.9) and (17.11), it can be
seen that there is no difference in whether the spring
is horizontal or vertical.

Problem 1. A mass of 1.5 kg is attached to

a vertical spring, as shown in Figure 17.4.

When the mass is displaced downwards a

distance of 55 mm from its position of rest,

it is observed to oscillate 60 times in

72 seconds. Determine (a) periodic time,

(b) the stiffness of the spring, (b) the time

taken to travel upwards a distance of 25 mm

for the first time, (c) the velocity at this

point.

q

q

30 mm

25 mm C

r= 55 m

vc,ac

vAsinq

vA

vAcosq

Static o

Figure 17.4

(a) Periodic time,

T=

72 seconds

60 oscillations

= 1 .2 seconds

(b) From equation (17.10),

T= 2 π

√

m

k

i.e. 1. 2 = 2 π

√

1. 5

k

Hence, 1. 22 =( 2 π)^2 ×

1. 5

k

from which, k=( 2 π)^2 ×

1. 5

1. 22

i.e. stiffness of spring, k= 41 .1N/m

(c) From Figure 17.4,

cosθ=

( 55 − 25 )

55

= 0. 545

from which, θ=cos−^10. 545 = 56. 94 °

Now, ω=

2 π

T

=

2 π

1. 2

= 5 .236 rad/s

But θ=ωt,

hence, timettaken to travel upwards a distance

of 25 mm, is given by:

t=

θ

ω

=

56. 94 °

5. 236

rad

s

×

2 πrad

360 °

= 0 .19 s

(d) Velocity atCin Figure 17.4,

vC=vAsinθ

=ωrsinθ

= 5. 236

rad

s

×

55

1000

m×sin 56. 94 °

= 0. 288 × 0 .838 m/s

i.e. vC= 0 .241 m/safter 25 mm of travel

Now try the following exercise

Exercise 85 Further problems on simple

harmonic motion

1. A particle oscillates 50 times in 22 s.
   Determine the periodic time and
   frequency.
   [T= 0 .44 s,f= 2 .27 Hz]


2. A yacht floats at a depth of 2.2 m. On
   a particular day, at a time of 09.30 h, the
   depth at low tide is 1.8 m and at a time of
   17.30 h, the depth of water at high tide is