PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-APPLICATION OF S.H.M
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PHYSICS FORM FIVE-SIMPLE HARMONIC MOTION-APPLICATION OF S.H.M
APPLICATION OF S.H.M
We shall consider the following cases of S.H.M
i) Oscillation of a Loaded Spring
ii) Oscillation of a Simple Pendulum
iii Oscillation of a Liquid in a U – tube
iv) Oscillation of a Floating Cylinder
v) Body Dropped in a funnel along earth diameter
vi) Oscillation of a ball placed in the Neck of Chamber Containing air
Oscillations of a Loaded Spring
If load attached to a spring is pulled a little from its mean position and then released the load will execute S.H.M
We shall consider the following two cases
1. Vibrations of a Horizontal spring
2. Vibrations of a Vertical spring
VIBRATIONS OF A HORIZONTAL SPRING
Consider a block of mass M attached to one end of a horizontal spring whi9le the other end of the spring is fixed to a rigid support
Fig. 7
The Block is at rest but is free to move along a friction less horizontal surface
In figure 7 is displaced through a small distance x to the right, the spring gets stretched
Fig.8
According to Hooke’s law, the spring exerts a restoring force F to the left given by
F = 
………………………………. (i)
Here k is the force constant (spring constant) and 
is the displacement of mass m from the mean position.
Clearly equation (i) satisfies the condition to produce S.H.M
If the block is released from the displaced position and left , the block will execute S.H.M
The time period (T) and frequency (f) of the vibrations can be obtained from
F = 
Ma = 
= 
From
2. VIBRATION OF A VERTICAL SPRING
Consider unloaded vertical spring of spring constant k
Fig. 9
Suppose the spring is loaded with a body of mass m and extended from its original length to an extension ‘e’
By Hookes law
mg = 
Now suppose the load is displaced down to distance x and then released. The applied force is given by
F = 
When realized the applied force is opposed by gravity force (weight)
Net result force = F – W
Ma = 
Ma = 
Ma = 
a = 
2 = 
From
The period of oscillation depends on mass of the loaded body and the spring constant.
In many practical situation springs are connected in series as well as in parallel.
SERIES AND PARALLEL CONNECTION OF SPRINGS
1. PARALLEL
Consider two springs of spring constant K1 and K2 arranged in parallel and then both loaded with a body of mass m as shown in fig. 10
Fig. 10
Suppose this body is displaced from its equilibrium position and the extension is x for the system to remain horizontal
Let F1 and F2 be the restoring forces acting on the springs.
Total force = F1+ F2
ma = -K1 
+ (-K2 
)
ma = 
Thus, the springs will execute S.H.M
From
TP = Periodic time for parallel connection
If 
(for identical spring)
From
OR
For
(for identical spring)
Alternative arrangment of parallel connection formula
Therefore if two identical sprinngs are arranged in parallel , their frequency increases by the factor of 
; since
is the frequency for a single spring for n-identical springs their frequency increases by a factor of 
.
Consider two springs S1and S2 of force constants k1 and k2 attached to a mass m and two fixed supports as shown in figure 11.
When the mass pulled downward, then length of the spring S1 will be extended by x while that of spring S2 will be compressed by x
Since the force constants of the two springs are different the restoring force exerted by each spring
Let F1 and F2 be the restoring forces exerted by springs S1 and S2 respectively
Both the restoring forces will be directed upward (opposite to displacement)
The resultant restoring force F
Fig .11
Effectively force constant of the system
2. SERIES CONNECTION
Consider two springs S1 and S2 of force constant K1, and K2 connected in series as shown below
Fig. 12
If x is the total displacement
Where x1 is extension due to spring of spring constant k1 and x2 to that of k2
F1 = 
F2 = 
F1 = F2 = F
The effective spring constant 
= periodic time for series connection.
For identical spring
Also since,
