Classical Mechanics
Classical Mechanics
5th Edition
ISBN: 9781891389221
Author: John R. Taylor
Publisher: University Science Books
Question
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Chapter 5, Problem 5.19P
To determine

Show that when the mass is displaced to a position (x,y) the potential energy is in the form of U=12k'r2 for an anisotropic oscillator. And also determine the constant k' in terms of k and the corresponding force.

Expert Solution & Answer
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Answer to Problem 5.19P

It is proved that when the mass is displaced to a position (x,y) the potential energy is in the form of U=12k'r2 for an anisotropic oscillator. And the constant k' in terms of k is k'=2k(2l0a) and the corresponding force is F=2k(2l0a)(xx^+yy^) .

Explanation of Solution

The four spring-mass system is

Classical Mechanics, Chapter 5, Problem 5.19P , additional homework tip  1

The mass is displaced a small distance to the position (x,y),

Classical Mechanics, Chapter 5, Problem 5.19P , additional homework tip  2

Write the expression for the potential energy of the spring

    U=12kx2        (I)

Here, U is the potential energy, k is the spring constant and x is the stretched or compressed length of the spring.

The length of spring 1 is

d1=(ay)2+x2        (II)

The length of spring 2 is

d2=(ax)2+y2        (III)

The length of spring 3 is

d3=(a+y)2+x2        (IV)

The length of spring 4 is

d4=(a+x)2+y2        (V)

Given, the mass is displaced to a small distance (x,y), xa and ya. Thus,

xa1  and ya1

Write the binomial expansion when t is small

    (1+t)12=1+12t18t2+...

Rewriting equation (II)

d1=[(ay)2+x2]12=[a2(1ya)2+x2]12=a[(1ya)2+x2a2]12=a[(12ya+y2a2)+x2a2]12

Using binomial expansion,

d1=a[1+(2ya+y2a2+x2a2)]12=a[1+12(2ya+y2a2+x2a2)18(2ya+y2a2+x2a2)2+...]

Given that x and y are small. Thus, neglecting the higher order powers for x and y in the above equation.

d1a[1+12(2ya+y2a2+x2a2)18(4y2a2)]a[1ya+12(y2a2+x2a2)18(4y2a2)]a[1ya+y22a2+12x2a2y22a2]a[1ya+12x2a2]

d1ay+12x2a

The original length of the spring is l0. Thus, the change in the length of the spring is

x1=d1l0=ay+12x2al0=(al0)y+12x2a

The potential energy of spring 1 is

U1=12kx12=12k[(al0)y+12x2a]2=12k[(al0)22y(al0)+y2+2(al0)(12x2a)+(12x2a)22y12x2a]=12k[(al0)22y(al0)+y2+(al0)(x2a)]

Given that x and y are small. Thus, neglecting the higher order powers for x and y in the above equation.

U112k[y22y(al0)+x2(1l0a)]+12k(al0)2+constant

U112k[y22y(al0)+x2(1l0a)]+constant        (VI)

For U2 , change yx and xy in equation (VI)

U212k[x22x(al0)+y2(1l0a)]+constant        (VII)

For U3 , change yy and xx in equation (VI)

U312k[y2+2y(al0)+x2(1l0a)]+constant        (VIII)

For U4 , change yx and xy in equation (VI)

U412k[x2+2x(al0)+y2(1l0a)]+constant        (IX)

The total potential energy of the system is

  U=U1+U2+U3+U4

Substitute equation (VI), (VII), (VIII) and (IX) in the above equation to solve for U

U=12k[y22y(al0)+x2(1l0a)]+constant+12k[x22x(al0)+y2(1l0a)]+constant+12k[y2+2y(al0)+x2(1l0a)]+constant+12k[x2+2x(al0)+y2(1l0a)]+constant=12k[2y2+2x2+x2(1l0a)+2y2(1l0a)]+constant=k[(x2+y2)+(x2+y2)(1l0a)]+constant=k(x2+y2)[1+(1l0a)]+constant U=k(x2+y2)[2l0a]+constant        (X)

Since, r2=x2+y2 and consider k'=2k(2l0a)

Equation (X) becomes,

U=12k'r2+constant

Hence it is proved that when the mass is displaced to a position (x,y) the potential energy is in the form of U=12k'r2 for an anisotropic oscillator.

Write the expression for force

    F=U

F=(Uxx^+Uyy^)

Ux=2xk(2l0a)

And,

Uy=2yk(2l0a)

Therefore,

F=(2xk(2l0a)x^+2yk(2l0a)y^)=2k(2l0a)(xx^+yy^)

Conclusion:

Hence it is proved that when the mass is displaced to a position (x,y) the potential energy is in the form of U=12k'r2 for an anisotropic oscillator. And the constant k' in terms of k is k'=2k(2l0a) and the corresponding force is F=2k(2l0a)(xx^+yy^) .

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