Engineering Mechanics

(^476) „„„„„ A Textbook of Engineering Mechanics
EXERCISE 23.1

1. A particle, moving with simple harmonic motion, has an acceleration of 6 m/s^2 at a distance
   of 1.5 m from the centre of oscillation. Find the time period of the oscillation.
   (Ans. 3.142 s)


2. A body weighing 150 N, moves with simple harmonic motion. The velocity and acceleration
   of the body when it is 200 mm from the centre of oscillation, are 5 m/s and 20 m/s^2 respectively.
   Determine (a) amplitude of motion and (b) no. of vibrations per minute.
   (Ans. 539 mm ; 95.5)


3. A particle moves with simple harmonic motion. When the particle is 0.75 m from the mid-
   path, its velocity is 11 m/s and when 2 m from the mid-path its velocity is 3 m/s. Find the
   angular velocity, periodic time and its maximum acceleration.
   (Ans. 5.7 rad/s ; 1.1 s ; 67.25 m/s^2 )


4. A particle moving with simple harmonic motion, has a velocity of 20 m/s at its central
   position. If the particle makes two oscillations per second, find (i) amplitude of motion
   and (ii) velocity at 1/4th the distance of the amplitude. (Ans. 1.59 m ; 19.35 m/s)

23.5. MAXIMUM VELOCITY AND ACCELERATION OF A PARTICLE

MOVING WITH SIMPLE HARMONIC MOTION

We have already discussed in Art. 23.4, that the velocity of a

particle moving with simple harmonic motion,

vry=ω^22 – ...(i)

A little consideration will show, that the velocity is maximum,

when y = 0 or when N passes through O i.e. its mean position. Therefore,

maximum velocity

vmax = ωr ...(ii)

It may be noted from equation (i) that its velocity is zero when y

= r, i.e. when N passes through Y' or Y as shown in Fig. 23.2. At these

points, N is momentarily at rest. We have also discussed that the

acceleration of a particle moving with simple harmonic motion,

a = ω^2 y ...(iii)

A little consideration will show, that the acceleration is maximum

when the value of y is maximum or y = r i.e. when N passes through Y or

Y'. Therefore maximum acceleration,

amax = ω^2 r ...(iv)

It may also be noted from equation (iii) that the acceleration is

zero, when y = 0 or when N passes through O i.e. its mean position. It is

thus obvious, that the acceleration is proportional to the distance from

O, i.e, mean position.

Example 23.6. A body is vibrating with simple harmonic motion of amplitude 100 mm, and

frequency 2 vibrations/sec. Calculate the maximum velocity and acceleration.

Solution. Given : Amplitude (r) = 100 mm = 0.1 m and frequency of body (N ) = 2 vib/sec.

Maximum velocity

We know that angular velocity of the body,

ω = 2πN = 2π × 2 = 4π rad/s

Simple pendulum is the most

common example for SHM.