From Newton’s second law, we can write the equation for the particle executing simple harmonic motion

**LINEAR SIMPLE HARMONIC OSCILLATOR (LHO)**

Consider a system containing a block of mass *m* attached to a massless spring with
stiffness constant or force constant or spring constant *k* placed on a smooth horizontal surface (frictionless surface) as
shown in Figure 10.13. Let *x*_{0}
be the equilibrium position or mean position of mass m when it is left
undisturbed. Suppose the mass is displaced through a small displacement *x* towards right from its equilibrium
position and then released, it will oscillate back and forth about its mean
position *x*_{0}. Let *F *be the restoring force (due to
stretching* *of the spring) which is
proportional to the amount of displacement of block. For one dimensional
motion, mathematically, we have

where negative sign implies that the restoring force
will always act opposite to the direction of the displacement. This equation is
called Hooke’s law. Notice that, the restoring force is linear with the
displacement (i.e., the exponent of force and displacement are unity). This is
not always true; in case if we apply a very large stretching force, then the
amplitude of oscillations becomes very large (which means, force is
proportional to displacement containing higher powers of x) and therefore, the
oscillation of the system is not linear and hence, it is called non-linear
oscillation. We restrict ourselves only to linear oscillations throughout our
discussions, which means Hooke’s law is valid (force and displacement have a
linear relationship).

From Newton’s second law, we can write the equation
for the particle executing simple harmonic motion

Comparing the equation (10.21) with simple harmonic
motion equation (10.10), we get

which means the angular frequency or natural frequency
of the oscillator is

Notice
that *in
simple harmonic motion, the*** time period of oscillation is independent of
amplitude**. This is valid only if the amplitude

The
solution of the differential equation of a SHM may be written as

where *A*, ω
and ϕ are constants. General solution for differential equation 10.21 is *x*(*t*)
=* A *sin(ω*t *+φ)+* B *cos(ω*t *+φ) where* *A and B are contants.

Let us consider a massless spring with stiffness
constant or force constant *k* attached
to a ceiling as shown in Figure 10.15. Let the length of the spring before
loading mass *m* be *L*. If the block of mass *m* is attached to the other end of
spring, then the spring elongates by a length *l*. Let F_{1} be the restoring force due to stretching of
spring. Due to mass *m*, the
gravitational force acts vertically downward. We can draw free-body diagram for
this system as shown in Figure 10.15. When the system is under equilibrium,

But the spring elongates by small displacement *l*, therefore,

Substituting equation (10.28) in equation (10.27), we
get

Suppose we apply a very small external force on the
mass such that the mass further displaces downward by a displacement *y*, then it will oscillate up and down.
Now, the restoring force due to this stretching of spring (total extension of
spring is *y* + *l* ) is

Since, the mass moves up and down with acceleration d^{2}y/dt^{2},
by drawing the free body, diagram for this case, we get

The net force acting on the mass due to this
stretching is

The
gravitational force opposes the restoring force. Substituting equation (10.29)
in equation (10.32), we get

*F *= −*ky *−* kl *+*
kl *= −*ky*

Applying
Newton’s law, we get

The above equation is in the form of simple harmonic
differential equation. Therefore, we get the time period as

The time period can be rewritten using equation
(10.29)

The acceleration due to gravity *g* can be computed from the formula

A
spring balance has a scale which ranges from 0 to 25 kg and the length of the
scale is 0.25m. It is taken to an unknown planet X where the acceleration due
to gravity is 11.5 m s^{−1}. Suppose a body of mass M kg is suspended
in this spring and made to oscillate with a period of 0.50 s. Compute the
gravitational force acting on the body.

Let us first calculate the stiffness constant of the
spring balance by using equation (10.29),

The time period of oscillations is given by T=2π√M/√*k*, , where M is the mass of the body.

Since, M is unknown, rearranging, we get

The gravitational force acting on the body is W = Mg =
7.3 × 11.5 = 83.95 N ≈ 84 N

Spring
constant or force constant, also called as stiffness constant, is a measure of
the stiffness of the spring. Larger the value of the spring constant, stiffer
is the spring. This implies that we need to apply more force to compress or
elongate the spring. Similarly, smaller the value of spring constant, the
spring can be stretched (elongated) or compressed with lesser force. Springs
can be connected in two ways. Either the springs can be connected end to end,
also known as series connection, or alternatively, connected in parallel. In
the following subsection, we compute the effective spring constant when

a.
Springs are connected in series

b.
Springs are connected in parallel

When two or more springs are connected in series, we
can replace (by removing) all the springs in series with an equivalent spring
(effective spring) whose net effect is the same as if all the springs are in
series connection. Given the value of individual spring constants *k*_{1},* k*_{2},* k*_{3},...
(known quantity), we can* *establish a
mathematical relationship to find out an effective (or equivalent) spring
constant *k _{s}* (unknown
quantity). For simplicity, let us consider only two springs whose spring
constant are

Let *F* be the
applied force towards right as shown in Figure 10.18. Since the spring
constants for different spring are different and the connection points between
them is not rigidly fixed, the strings can stretch in different lengths. Let *x*_{1} and *x*_{2} be the elongation of springs from their equilibrium
position (un-stretched position) due to the applied force *F*. Then, the net displacement of the mass point is

From Hooke’s law, the net force

For springs in series connection

−*k*1*x*1 = −*k*2*x*2 = *F*

Therefore, substituting equation (10.39) in equation
(10.38), the *effective spring constant*
can be calculated as

Suppose we have *n*
springs connected in series, the effective spring constant in series is

If all spring constants are identical i.e., *k*_{1} = *k*_{2}* *=... =* k*_{n}* *=* k *then

This means that the effective spring constant reduces
by the factor *n*. Hence, for springs in series connection, the effective
spring constant is lesser than the individual spring constants.

From
equation (10.39), we have,

*k*_{1}*x*_{1}* *=* k*_{2}*x*_{2}

Then
the ratio of compressed distance or elongated distance *x*_{1} and *x*_{2}
is

The
elastic potential energy stored in first and second springs are V_{1}=1/2
k_{1}x_{1}^{2}_{ }and V_{2}=1/2 k_{2}x_{2}^{2}_{
}respectively. Then, their ratio is

Consider two springs whose force constants are 1 N m^{−1}
and 2 N m^{−1} which are connected in series. Calculate the effective
spring constant (*k _{s}* ) and
comment on

*k _{s} *<

Therefore,
the effective spring constant is lesser than both *k*_{1} and *k*_{2}.

When two or more springs are connected in parallel, we
can replace (by removing) all these springs with an equivalent spring
(effective spring) whose net effect is same as if all the springs are in
parallel connection. Given the values of individual spring constants to be *k*_{1},*k*_{2},*k*_{3},
... (known quantities), we can establish a mathematical relationship to find out
an effective (or equivalent) spring constant *k*_{p}* *(unknown
quantity). For simplicity, let us* *consider
only two springs of spring constants *k*_{1}and* k*_{2}* *attached to a mass* m *as
shown in* *Figure 10.19. The results
can be generalized to any number of springs in parallel.

Let the force *F*
be applied towards right as shown in Figure 10.20. In this case, both the
springs elongate or compress by the same amount of displacement. Therefore, net
force for the displacement of mass *m*
is

where *k*_{p}
is called ** effective spring constant**. Let the first spring be elongated by
a displacement

Equating equations (10.46) and (10.45), we get

Generalizing, for n springs connected in parallel,

If all spring constants are identical i.e., *k*_{1}* *=* k*_{2}= ... =* k _{n} *=

This implies that the effective spring constant
increases by a factor *n*. Hence, for
the springs in parallel connection, the effective spring constant is greater
than individual spring constant.

Consider
two springs with force constants 1 N m^{−1} and 2 N m^{−1}
connected in parallel. Calculate the effective spring constant (*k _{p}* ) and comment on

*k _{1} *= 1 N m

*k _{p} *=

*k _{p} *= 1 + 2 = 3 N m

*k _{p} *>

Therefore,
the effective spring constant is greater than both *k*_{1} and *k*_{2}.

Calculate the equivalent spring constant for the
following systems and also compute if all the spring constants are equal:

a.
Since *k*_{1} and *k*_{2} are parallel, *k*_{u} = *k*_{1} + *k*_{2}
Similarly, *k*_{3} and *k*_{4} are parallel, therefore, *k*_{d} = *k*_{3} + *k*_{4}

But
*k*_{u} and *k*_{d} are in series,

If
all the spring constants are equal then, *k*_{1}
= *k*_{2} = *k*_{3} = *k*_{4} = *k*

Which
means, *k*_{u} = 2*k* and *k*_{d} = 2*k*

b.
Since *k*_{1} and *k*_{2} are parallel, *k*_{A} = *k*_{1} + *k*_{2}
Similarly, *k*_{4} and *k*_{5} are parallel, therefore, *k*_{B} = *k*_{4} + *k*_{5}

But
*k*_{A}, *k*_{3}, *k*_{B},
and *k*_{6} are in series,

If all the spring constants are equal then, *k*_{1} = *k*_{2} = *k*_{3}
= *k*_{4} = *k*_{5} = *k*_{6} = *k* which
means, *k*_{A} = 2*k* and *k*_{B} = 2*k*

*k _{eq}
*= k/3

A
mass *m* moves with a speed *v* on a horizontal smooth surface and
collides with a nearly massless spring whose spring constant is *k*. If the mass stops after collision,
compute the maximum compression of the spring.

When
the mass collides with the spring, from the law of conservation of energy “the
loss in kinetic energy of mass is gain in elastic potential energy by spring”.

Let
*x* be the distance of compression of
spring, then the law of conservation of energy

A
pendulum is a mechanical system which exhibits periodic motion. It has a bob
with mass m suspended by a long string (assumed to be massless and inextensible
string) and the other end is fixed on a stand as shown in Figure 10.21 (a). At
equilibrium, the pendulum does not oscillate and hangs vertically downward.
Such a position is known as mean position or equilibrium position. When a
pendulum is displaced through a small displacement from its equilibrium position
and released, the bob of the pendulum executes to and fro motion. Let *l* be the length of the pendulum which is
taken as the distance between the point of suspension and the centre of gravity
of the bob. Two forces act on the bob of the pendulum at any displaced
position, as shown in the Figure 10.21 (d),

(i) The gravitational force acting on the body ( ) which acts vertically downwards.

(ii) The tension in the string which acts along the string to the point of suspension.

Resolving
the gravitational force into its components:

**a. Normal component: **The component** **along the string but in opposition to the direction of tension, *F _{as}* =

**b. Tangential component: **The** **component perpendicular to the string i.e., along tangential
direction of arc of swing, *F _{ps}* =

Therefore,
The normal component of the force is, along the string,

From the Figure 10.21, we can observe that the
tangential component *W _{ps}*
of the gravitational force always points towards the equilibrium position i.e.,
the direction in which it always points opposite to the direction of
displacement of the bob from the mean position. Hence, in this case, the
tangential force is nothing but the restoring force. Applying Newton’s second
law along tangential direction, we have

where, *s* is
the position of bob which is measured along the arc. Expressing arc length in
terms of angular displacement i.e.,

Substituting equation (10.53) in equation (10.51), we
get

Because of the presence of sin *θ* in the above differential equation, it is a non-linear
differential equation (Here, homogeneous second order). Assume “the small
oscillation approximation”, sin *θ* ≈ *θ*, the above differential equation
becomes linear differential equation.

This is the well known oscillatory differential
equation. Therefore, the angular frequency of this oscillator (natural
frequency of this system) is

The
time period of a simple pendulum

For
a given value of acceleration due to gravity, the time period of a simple
pendulum is directly proportional to the square root of length of the pendulum.

For a fixed length, the time period of a simple
pendulum is inversely proportional to square root of acceleration due to
gravity.

The
time period of oscillation is independent of mass of the simple pendulum.
This is similar to free fall. Therefore, in a pendulum of fixed length, it does
not matter whether an elephant swings or an ant swings. Both of them will swing
with the same time period.

For a pendulum with small angle approximation (angular
displacement is very small), the time period is independent of amplitude of the
oscillation.

In simple pendulum experiment, we have used small
angle approximation . Discuss the small angle approximation.

For θ in radian, sin θ ≈ θ for very small angles

This
means that “for *θ* as large as 10
degrees, sin *θ* is nearly the same as *θ* when *θ* is expressed in radians”.
As *θ* increases in value sin*θ* gradually becomes different from *θ*

Suppose
the suspended wire is affected due to change in temperature. The rise in
temperature affects length by

*l *=* l*_{o}* *(1 +*
α *∆t)

where
*l*_{o} is the original length
of the wire and *l* is final length of
the wire when the temperature is raised. Let ∆*t* is the change in temperature and *α* is the co-efficient of linear expansion.

where ∆T is the change in time period due to the
effect of temperature and *T*_{0}
is the time period of the simple pendulum with original length *l*_{0}.

If
the length of the simple pendulum is increased by 44% from its original length,
calculate the percentage increase in time period of the pendulum.

**Oscillation
of liquid in a U-tube:**

Consider a U-shaped glass tube which consists of two
open arms with uniform cross-sectional area A. Let us pour a non-viscous
uniform incompressible liquid of density ρ in the U-shaped tube to a *h*eight *h* as shown in the Figure 10.22. If the liquid and tube are not
disturbed then the liquid surface will be in equilibrium position *O.* It means the pressure as measured at
any point on the liquid is the same and also at the surface on the arm (edge of
the tube on either side), which balances with the atmospheric pressure. Due to
this the level of liquid in each arm will be the same. By blowing air one can
provide sufficient force in one arm, and the liquid gets disturbed from
equilibrium position *O*, which means,
the pressure at blown arm is higher than the other arm. This creates difference
in pressure which will cause the liquid to oscillate for a very short duration
of time about the mean or equilibrium position and finally comes to rest.

Time period of the oscillation is

Tags : Oscillations | Physics , 11th Physics : UNIT 10 : Oscillations

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

11th Physics : UNIT 10 : Oscillations : Linear Simple Harmonic Oscillator (LHO) | Oscillations | Physics

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2024 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.