Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 15, Problem 46P

(a)

To determine

To calculate:

C/A=2v2/(v1+v2) and B/A=(v1+v2)/(v1+v2)

(a)

Expert Solution
Check Mark

Answer to Problem 46P

It is proved that B/A is equals to the v1-v2/v1+v2 and C/A is equals to the 2v2/v1+v2 .

Explanation of Solution

Given:

The speed of the wave is v1 when region x<0 .

The speed of the wave is v2 when region x>0 .

Formula used:

Consideration: limB/A|v2/v10

This limit gives B/A=+1 , that is telling that the transmitted wave has standing wave with node at x=0 .

Calculation:

Both the wave function and its first spatial derivative are continuous at x=0 to establish equations relating A,B,C and k1,k2 .

Let, the y1(x,t) represents the wave function in the region x<0 and y2(x,t) represents the wave function in the region x>0

Now, to expression for continuity of the two-wave function’s at x=0 are,

  y(0,t)=y2(0,t)

And

  Asin[k1(0)-ωt]+Bsin[k1(0)+ωt]=Csin[k2(0)-ωt]

Or

  Asin(-ωt)+Bsinωt=Csin(-ωt)

The sine function is odd as it is symmetric about the origin. So, Sin()=-sinθ .

  -Asinωt+Bsinωt=-Csinωt and A-B=C ……(1)

Now, differentiate the wave functions with respect to x to get,

  y1x=Ak1cos(k1,x-ωt)+Bk1cos(k1,x+ωt)

And

  y2x=Ck2cos(k2,x-ωt)

Now, express the continuity of the slopes of the two wave functions at x=0 ,

   y 1x|x=0= y 2x|x=0

And

  Ak1cos[k1(0)-ωt]+Bk1cosωt=Ck2cos(-ωt)

The cosine function is even as it is symmetric about the y -axis. So, cos()=cosθ .

  Ak1cosωt+Bk1cosωt=Ck2cosωt and k1A+k1B=k2C …(2)

Multiply the equation (1) by k1 and add it to equation (2) to get,

  2k1A=(k1+k2)C .

Now to solving for,

  C=2k1k1+k2A=21+k2/k1A

Solve for C/A and substitute ω/v1 for k1 and ω/v2 for k2 to get,

  CA=2 1+k2 /k1CA=2 1+v1 /v2CA= 2v2v1 +v2

Now, put the equation (1) to get,

  A-B=( 2v2v2 +v1)A

Solving for B/A yields,

  BA=v1-v2v1+v2

Conclusion:

Hence, it is proved BA=v1-v2v1+v2 and CA=2v2v1+v2 .

(b)

To determine

To calculate: B2+(v1+v2)C2=A2

(b)

Expert Solution
Check Mark

Answer to Problem 46P

The identified equation is,

  B2+(v1v2)C2=A2

Explanation of Solution

Given:

The speed of the wave is v1 when region x<0 .

The speed of the wave is v2 when region x>0 .

Formula used:

Consideration: limB/A|v2/v10

This limit gives B/A=+1 , that is telling that the transmitted wave has standing wave with node at x=0 .

Calculation:

To show the B2+(v1+v2)C2=A2 , there is need to use results of the part (a) to get,

  B=-[(1-α)/(1+α)]A and C=2A/(1+α)

In which,

  α=v1/v2

After substituting the values in equation,

  B2+(v1+v2)C2=A2

And then check to see if result equation is identified or not,

  B2+(v1 +v2)C2=A2( 1-α 1+α)2A2( 2 1+α)2A2=A2( 1-α 1+α)2( 2 1+α)2=1 ( 1-α )2+4α ( 1+α )2=1 1-2α+α2+4α ( 1+α )2=1 1-2α+α2 ( 1+α )2=1 ( 1+α )2 ( 1+α )2=11=1

Therefore, identified equation is,

  B2+(v1v2)C2=A2

Conclusion:

Therefore, identified equation is B2+(v1v2)C2=A2

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Chapter 15 Solutions

Physics for Scientists and Engineers

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