Physics, asked by ashpak263gmailcom, 5 months ago

derive the expression for energy in S.H.M show the variations of energy with displacement​

Answers

Answered by KhadGS123
1

Answer:

To produce a deformation in an object, we must do work. That is, whether you pluck a guitar string or compress a car’s shock absorber, a force must be exerted through a distance. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy.

Consider the example of a block attached to a spring on a frictionless table, oscillating in SHM. The force of the spring is a conservative force (which you studied in the chapter on potential energy and conservation of energy), and we can define a potential energy for it. This potential energy is the energy stored in the spring when the spring is extended or compressed. In this case, the block oscillates in one dimension with the force of the spring acting parallel to the motion:

W

=

x

f

x

i

F

x

d

x

=

x

f

x

i

k

x

d

x

=

[

1

2

k

x

2

]

x

f

x

i

=

[

1

2

k

x

2

f

1

2

k

x

2

i

]

=

[

U

f

U

i

]

=

Δ

U

.

When considering the energy stored in a spring, the equilibrium position, marked as

x

i

=

0.00

m,

is the position at which the energy stored in the spring is equal to zero. When the spring is stretched or compressed a distance x, the potential energy stored in the spring is

Answered by bidyutsingh6
0

Answer:

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Lumen

Physics

Oscillatory Motion and Waves

Energy and the Simple Harmonic Oscillator

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Determine the maximum speed of an oscillating system.

To study the energy of a simple harmonic oscillator, we first consider all the forms of energy it can have We know from Hooke’s Law: Stress and Strain Revisited that the energy stored in the deformation of a simple harmonic oscillator is a form of potential energy given by:

PE

el

=

1

2

k

x

2

.

Because a simple harmonic oscillator has no dissipative forces, the other important form of energy is kinetic energy KE. Conservation of energy for these two forms is:

KE + PEel = constant

or

1

2

m

v

2

+

1

2

k

x

2

=

constant

.

This statement of conservation of energy is valid for all simple harmonic oscillators, including ones where the gravitational force plays a role

Namely, for a simple pendulum we replace the velocity with v = Lω, the spring constant with

k

=

m

g

L

, and the displacement term with x = Lθ. Thus

1

2

m

L

2

ω

2

+

1

2

m

g

L

θ

2

=

constant

In the case of undamped simple harmonic motion, the energy oscillates back and forth between kinetic and potential, going completely from one to the other as the system oscillates. So for the simple example of an object on a frictionless surface attached to a spring, as shown again in Figure 1, the motion starts with all of the energy stored in the spring. As the object starts to move, the elastic potential energy is converted to kinetic energy, becoming entirely kinetic energy at the equilibrium position. It is then converted back into elastic potential energy by the spring, the velocity becomes zero when the kinetic energy is completely converted, and so on. This concept provides extra insight here and in later applications of simple harmonic motion, such as alternating current circuits.

Figure a shows a spring on a frictionless surface attached to a bar or wall from the left side, and on the right side of it there’s an object attached to it with mass m, its amplitude is given by X, and x equal to zero at the equilibrium level. Force F is applied to it from the right side, shown with left direction pointed red arrow and velocity v is equal to zero. A direction point showing the north and west direction is also given alongside this figure as well as with other four figures. The energy given here for the object is given according to the velocity. In figure b, after the force has been applied, the object moves to the left compressing the spring a bit, and the displaced area of the object from its initial point is shown in sketched dots. F is equal to zero and the V is max in negative direction. The energy given here for the object is given according to the velocity. In figure c, the spring has been compressed to the maximum level, and the amplitude is negative x. Now the direction of force changes to the rightward direction, shown with right direction pointed red arrow and the velocity v zero. The energy given here for the object is given according to the velocity. In figure d, the spring is shown released from the compressed level and the object has moved toward the right side up to the equilibrium level. F is zero, and the velocity v is maximum. The energy given here for the object is given according to the velocity. In figure e, the spring has been stretched loose to the maximum level and the object has moved to the far right. Now again the velocity here is equal to zero and the direction of force again is to the left hand side, shown here as F is equal to zero. The energy given here for the object is given according to the velocity.

Figure 1. The transformation of energy in simple harmonic motion is illustrated for an object attached to a spring on a frictionless surface.

The conservation of energy principle can be used to derive an expression for velocity v. If we start our simple harmonic motion with zero velocity and maximum displacement (x = X), then the total energy is

1

2

k

X

2

.

This total energy is constant and is shifted back and forth between kinetic energy and potential energy, at most times being shared by each. The conservation of energy for this system in equation form is thus:

1

2

m

v

2

+

1

2

k

x

2

=

1

2

k

X

2

.

Solving this equation for v yields:

v

=

±

k

m

(

X

2

x

2

)

.

Manipulating this expression algebraically gives:

v

=

±

k

m

X

1

x

2

X

2

and so

v

=

±

v

max

1

x

2

X

2

,

where

v

max

=

k

m

X

.

From this expression, we see that the velocity is a maximum (vmax) at x = 0, as stated earlier in

v

(

t

)

=

v

max

sin

2

π

t

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