Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
Question
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Chapter 15, Problem 36P
To determine

Find the correct options which satisfy the wave equation:

    (a) y(x,t)=k(x+vt)3

    (b) y(x,t)=Aeik(x-vt)

    (c) y(x,t)=ln[k(x-vt)]

Expert Solution & Answer
Check Mark

Answer to Problem 36P

Option (a), (b) and (c) are correct as it satisfies the general wave equation that is 2y/x22y/t2=1v2 .

Explanation of Solution

Given:

Equation:

  1. y(x,t)=k(x+vt)3 .
  2. y(x,t)=Aeik(x-vt)
  3. y(x,t)=ln[k(x-vt)]

Calculation:

To show the given function satisfies the wave equation.

The wave equation:

  2yx2=(1v2)2yt2

So, initially need to find their first and second derivatives with respect to x and

t and then substitute these derivatives in the wave equation.

The first two spatial derivatives of y(x,t)=k(x+vt)3 ,

  yx=3k(x+vt)2

And

  2yx2=6k(x+vt) …….(1)

Now, first two temporal derivatives of y(x,t)=k(x+vt)3 ,

  yt=3kv(x+vt)2

And

  2yt2=6kv2(x+vt) ……(2)

The ratio of the equation (1) to equation (2) is,

   2 y x 2 2 y t 2 =6k( x+vt) 6kv2( x+vt) 2 y x 2 2 y t 2 =1v2

Thus, option (a) is correct answer as it is confirming that y(x,t)=k(x+vt)3 satisfies the general wave equation.

Option (b) is correct answer as it is confirming that y(x,t)=Aeik(x-vt) satisfies the general wave equation.

Reason:

Find the first two spatial derivatives of y(x,t)=Aeik(x-vt) ,

  yx=ikAeik(x-vt) , 2yx2=i2k2Aeik(x-vt)

Or

  2yx2=-k2Aeik(x-vt) ……(1).

Now, first two temporal derivatives of y(x,t)=Aeik(x-vt) ,

  yt=-ikvAeik(x-vt) , 2yt2=i2k2Aeik(x-vt)

Or     2yt2=-k2v2Aeik(x-vt) ……(2).

The ratio of the equation (1) to equation (2) is,

   2 y x 2 2 y t 2 = -k2 Ae ik( x-vt ) -k2v2 Ae ik( x-vt ) 2 y x 2 2 y t 2 =1v2

Thus, confirming that y(x,t)=Aeik(x-vt) satisfies the general wave equation.

Option (c) is correct answer as it is confirming that y(x,t)=ln[k(x-vt)] satisfies the general wave equation.

Reason:

first two spatial derivatives of y(x,t)=ln[k(x-vt)] ,

  yx=kx-vt , 2yx2=k2( x-vt)2 …….(1)

Now, first two temporal derivatives of y(x,t)=ln[k(x-vt)] ,

  yt=vkx-vt , 2yt2=v2k2( x-vt)2 …….(2)

The ratio of the equation (1) to equation (2) is,

   2 y x 2 2 y t 2 = k 2 ( x-vt ) 2 v 2 k 2 ( x-vt ) 2 2 y x 2 2 y t 2 =1v2

Thus, confirming that y(x,t)=ln[k(x-vt)] satisfies the general wave equation.

Conclusion:

Thus, all three options satisfy the given general wave equation that is 2y/x22y/t2=1v2 .

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