(^474) „„„„„ A Textbook of Engineering Mechanics
We know that velocity of the particle,
vry=ω^22 –
∴ 9–(2) –4=ω rr^22 =ω^2 ...(i)
and 4–(3) –9=ω rr^22 =ω^2 ...(ii)
Dividing equation (i) by (ii)
2
2
9 –4
(^4) –9
r
r

or
2
2
81 – 4
(^16) –9
r
r
= ...(Squaring both sides)
81 r^2 – 729 = 16r^2 – 64 or 65 r^2 = 665
∴
665
10.23 3.2 m
65
r== =
Time-period of the particle
Substituting this value of r in equation (i),
9 =ω (3.2) – 4^2 =ω 6.24=ω×2.5
∴
9
3.6 rad/s
2.5
ω= =
We know that time period,
22
1.75 s
3.6
T
ππ
== =
ω
Ans.
Example 23.4. In a system, the amplitude of the motion is 5 m and the time is 4 seconds.
Find the time required by the particle in passing between points which are at distances of 4 m and 2
m from the centre of force and are on the same side of it.
Solution. Given : Amplitude (r) = 5 m ; Time taken (T) = 4 sec and distances of the point ( y 1 )
= 4 m and ( y 2 ) = 2m
We know that angular velocity of the particle,
22
90 / s
T 4
ππ
ω===°
and displacement of particle ( y),
4 = r sin ωt = 5 sin ωt 1
∴ sin 1 4 0.8
5
ω= =t
ωt 1 = 53.1°
or 1
53.1
0.59 s
90
t ==
Similarly 2 = r sin ωt 2
or 2
2
sin 0.4
5
ω= =t
∴ωt 2 = 23.6°
or 2
23.6
0.26 s
90
t ==
Fig. 23.4.