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K t | avg = 1 2 X 1 2 μ v ω 2 A 2 = 1 4 μ v ω 2 A 2

Elastic potential energy

The elastic potential energy of the string element results as string element is stretched during its oscillation. The extension or stretching is maximum at mean position. We can see in the figure that the length of string element of equal x-length “dx” is greater at mean position than at the extreme. As a matter of fact, the elongation depends on the slope of the curve. Greater the slope, greater is the elongation. The string has the least length when slope is zero. For illustration purpose, the curve is purposely drawn in such a manner that the elongation of string element at mean position is highlighted.

Elongation in the string

The string element stretched most at equilibrium position.

Greater extension of string element corresponds to greater elastic energy. As such, it is greatest at mean position and zero at extreme position. This deduction is contrary to the case of SHM in which potential energy is greatest at extreme positions and zero at mean position.

The elastic potential energy of an element of mass “dm” is given by the expression :

d U = 1 2 m v 2 y x 2

where “v” is wave speed and “ y x ” is the slope of waveform. The respective expressions for these quantities are :

v = ω k

y x = k A cos k x ω t

Putting in the expression of elastic potential energy,

d U = μ x ω 2 k 2 A 2 cos 2 k x ω t 2 k 2 = 1 2 μ x ω 2 A 2 cos 2 k x ω t

This is the same expression as derived for kinetic energy. The elastic potential energy per unit length is :

U L = U x = 1 2 μ ω 2 A 2 cos 2 k x ω t

Rate of transmission of elastic potential energy

The rate, at which elastic potential energy is transmitted, is obtained by dividing expression of kinetic energy by small time element, “dt”. This expression is same as that for kinetic energy.

U t = 1 2 μ v ω 2 A 2 cos 2 k x ω t

and average rate of transmission of elastic potential energy is :

U t | avg = 1 2 X 1 2 μ v ω 2 A 2 = 1 4 μ v ω 2 A 2

Mechanical energy

Since the expression of elastic potential energy is same as that of kinetic energy, we get mechanical energy expressions by multiplying expression of kinetic energy by “2”. The mechanical energy associated with small string element, "dx", is :

d E = 2 X d K = 2 X 1 2 m v p 2 = μ x ω 2 A 2 cos 2 k x ω t

Similarly, the mechanical energy per unit length is :

E L = 2 X 1 2 μ ω 2 A 2 cos 2 k x ω t = μ ω 2 A 2 cos 2 k x ω t

Rate of transmission of mechanical energy

Mechanical energy transmitted by the wave is sum of kinetic energy and elastic potential energy. The instantaneous rate of transmission of mechanical energy, therefore, is also obtained by multiplying instantaneous kinetic energy by “2”.:

E t = 2 X K t = μ v ω 2 A 2 cos 2 k x ω t

Energy is greatest at the mean position and zero at the extreme positions of the particle in SHM. How can this happen? We need to interpret this fact in the light of transmission of energy from one section of the string to another. When element position changes in transverse direction away from mean position, it passes on the energy to the neighboring element. In doing so, the energy of string element is reduced, whereas the particle following it in the direction of wave motion gains energy (we consider no energy loss in the process). In more direct words, we can say that elements pass on the disturbance to the adjacent element. As the element moves to the extreme position, it completely exhausts its energy and acquires its undisturbed length. However, it yields to next train of energy or disturbance as the free end of the string is moved again and again.

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Source:  OpenStax, Oscillation and wave motion. OpenStax CNX. Apr 19, 2008 Download for free at http://cnx.org/content/col10493/1.12
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