Engineering Mechanics

(^538) „„„„„ A Textbook of Engineering Mechanics
Fig. 26.6.
(d) Maximum angular velocity of the bob
We also know that maximum angular velocity of the bob,
0·27
0·174 rad/s
1·5 5
vmax
l
=== Ans.
Example 26.9. A simple pendulum consists of a 600 mm long cord and a bob of mass 2 kg.
Find the no. of oscillations made by the bob per second.
If the same pendulum is suspended inside a train, accelerating smoothly on a level track at
the rate of 3 m/s^2 , find the angle which the cord will make with the vertical. Also find the tension
in the cord.
Solution. Given: Length of cord (l) = 600 mm = 0·6 m ; Mass of bob (m) = 2 kg and
acceleration (a) = 3 m/s^2
Number of oscillations made by the bob per second
We know that time-period for one oscillation,
0·6
22 1·55s
9·8
l
t
g
=π =π =
and no. of oscillations made by the bob per second,
1
0·65 Hz
1·5 5
l
n
t
== = Ans.
Angle, which the cord will make with the vertical
Let θ = Angle, which the cord will make with the vertical.
We know that weight of the bob (acting downwards)
= mg = 2 × 9·8
= 19·6 N
and inertia force acting on the bob (opposite to the acceleration of the train)
= ma = 2 × 3
= 6 N
∴
6
tan 0·3061
19·6
θ= = or θ = 17°
Tension in the cord
We know that tension in the cord,
(^) T=+=(19·6)^22 (6) 20·5 N Ans.
26.6.GAIN OR LOSS IN THE NO. OF OSCILLATIONS DUE TO CHANGE IN THE
LENGTH OF STRING OR ACCELERATION DUE TO GRAVITY OF A SIMPLE
PENDULUM
Consider a simple pendulum, oscillating with a simple harmonic motion.
Let l = Length of the string,