Engineering Mechanics

Chapter 23 : Simple Harmonic Motion „„„„„ 473

∴ Velocity of piston, vry=ω^22 –4(1)–(0.75)8.3m/s= π^22 = Ans.

Acceleration of piston

We know that acceleration of piston,
a = ω^2 y = (4π)^2 × 0.75 = 118.4 m/s^2 Ans.
Example 23.2. A body, moving with simple harmonic motion, has an amplitude of 1 meter
and the period of complete oscillation is 2 seconds. What will be the velocity and acceleration of
the body after 0.4 second from the extreme position?

Solution. Given : Amplitude (r) = 1 m; Periodic time (T) = 2 s and time taken by the body
from extreme position = 0.4 s

Velocity of the body

Now let O be the centre, Y an extremity of the motion and P the position of the body after
0.4 sec from Y as shown in Fig. 23.3. Therefore time required by the body to travel from Y to P

= 0.4 s

We know that time required by the body to travel from O to Y

1

4

=×T

1

20.5s

4

=×=

∴Time required by the body to travel from O to N

t = 0.5 – 0.4 = 0.1 s ...(i)

We know that angular velocity of the body

22

rad/s

T 2

ππ

ω===π ...(ii)

∴ Displacement of the body after 0.4 sec from the extreme position (or 0.1 second from the
mean position),
y = r cos ωt = 1 cos (π × 0.1) = 1 cos 18° m
...(π × 0.1 = 180 × 0.1 = 18°)
= 0.95 m
∴ Velocity of the body,
vry=ω^22 – =π 1 – (0.95)^22 =0.98 m/s^ Ans.

Acceleration of the body
We know that acceleration of the body,
a = ω^2 y = (π)^2 × 0.95 = 9.38 m/s^2 Ans.
Example 23.3. Find amplitude and time period of a particle moving with simple harmonic
motion, which has a velocity of 9 m/s and 4 m/s at the distance of 2 m and 3 m respectively from
the centre.

Solution. Given : When velocity (v 1 ) = 9 m/s, distance from centre ( y 1 ) = 2 m and when
velocity (v 2 ) = 4 m/s, distance from centre ( y 2 ) = 3 m

Amplitude of the particle

Let r = Amplitude of the particle, and

ω = Angular velocity of the particle.

Fig. 23.3.