Engineering Mechanics

(^472) „„„„„ A Textbook of Engineering Mechanics
23.4.VELOCITY AND ACCELERATION OF A PARTICLE MOVING WITH
SIMPLE HARMONIC MOTION
Consider a particle moving along the circumference of a circle,
of radius r with a uniform angular velocity of ω radians/sec as shown
in Fig. 23.2.
Let P be the position of the particle at some instant after t sec
from X. Therefore, angle turned by the particle,
θ = ωt rad
and displacement of the point N (i.e. projection of P on the vertical
diameter Y-Y' of the circle),
y = ON = r sin θ
= r sin ωt ...(i)
Differentiating this equation with respect to time t,
cos
dy
rt
dt
=ω ω ...(ii)
or velocity, vr=ω1–sin^2 ωt ...(Q sin^2 θ + cos^2 θ = 1)
From equation (i), we find that
sin
y
t
r
ω=
Substituting this value of sin ωt in the above equation,
2
1– 2
y
r
r
⎛⎞
ν= ω ⎜⎟⎜⎟
⎝⎠
or velocity, vry=ω^22 – ...(iii)
Now differentiating equation (ii) with respect to time t,
2
2
2 –sin
dy
rt
dt
=ω ω
or acceleration, a = – ω^2 y ...(Substituting y = r sin ωt)
Note. The minus sign shows that the direction of acceleration is opposite to the direction in
which y increases i.e., the acceleration is always directed towards the point O. But in actual practice,
this relation is used as a = ω^2 y
Example 23.1. The piston of a steam engine moves with simple harmonic motion. The crank
rotates at 120 r.p.m. and the stroke length is 2 metres. Find the velocity and acceleration of the
piston, when it is at a distance of 0.75 metre from the centre.
Solution. Given : Frequency of piston (N) = 120 r.p.m ; stroke length l = 2 m or radius (r)
= 1 m and distance of piston from the centre (y) = 0.75 m
Velocity of piston
We know that angular velocity of piston,
2 2 120
4 rad/sec
60 60
ππ×N
ω= = = π
Fig. 23.2. Particle moving
with S.H.M.