Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 11, Problem 87P

(a)

To determine

The wave with higher wave speed and its value.

(a)

Expert Solution
Check Mark

Answer to Problem 87P

The wave with higher wave speed is y(x,t)=(1.50cm)sin[(4.00cm1)x+(6.00s1)t]_ and its speed is 1.50cm/s_.

Explanation of Solution

The given wave equations are,

y(x,t)=(1.50cm)sin[(4.00cm1)x+(6.00s1)t] (I)

y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t] (II)

Write the general expression for the travelling waves.

y(x,t)=Asin[kx+ωt] (III)

Here, A is the amplitude, k is the wave vector, and ω is the angular frequency.

Write the expression for the wave speed of the given waves (I) and (II).

vI=ωIkI (IV)

vII=ωIIkII (V)

Conclusion:

Substitute 6.00s1 for ωI, 4.00cm1 for kI in equation (IV) to find vI.

vI=6.00s14.00cm1=1.50cm/s

Substitute 3.00s1 for ωII, 3.00cm1 for kII in equation (V) to find vII.

vI=3.00s13.00cm1=1.00cm/s

It is clear from above that vI>vII.

Therefore, the wave with higher wave speed is y(x,t)=(1.50cm)sin[(4.00cm1)x+(6.00s1)t]_ and its speed is 1.50cm/s_.

(b)

To determine

The wave with higher wavelength and its value.

(b)

Expert Solution
Check Mark

Answer to Problem 87P

The wave with higher wavelength is y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t]_ and its wavelength is 2.09cm_.

Explanation of Solution

The given wave equations are,

y(x,t)=(1.50cm)sin[(4.00cm1)x+(6.00s1)t]

y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t]

Write the expression for the wavelength λ of the given waves (I) and (II).

λI=2πkI (VI)

λII=2πkII (VII)

Conclusion:

Substitute 4.00cm1 for kI in equation (VI) to find λI.

λI=2π4.00cm1=1.57cm

Substitute 3.00cm1 for kII in equation (VII) to find λII.

λI=2π3.00cm1=2.09cm

It is clear from above that λII>λI.

Therefore, the wave with higher wavelength is y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t]_ and its wavelength is 2.09cm_.

(c)

To determine

The wave with the faster maximum speed of a point in the medium and its value.

(c)

Expert Solution
Check Mark

Answer to Problem 87P

The wave with faster maximum speed of a point in the medium is y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t]_ and that speed is 13.5cm/s_.

Explanation of Solution

The given wave equations are,

y(x,t)=(1.50cm)sin[(4.00cm1)x+(6.00s1)t]

y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t]

Write the expression for the maximum speed vm of a point in the medium for each traveling wave..

vmI=ωIAI (VIII)

vmII=ωIIAII (IX)

Conclusion:

Substitute 6.00s1 for ωI, 1.50cm for AI in equation (VIII) to find vmI.

vmI=(6.00s1)(1.50cm)=9.00cm/s

Substitute 3.00s1 for ωII, 4.50cm for AI in equation (VIII) to find vmI.

vmII=(3.00s1)(4.50cm)=13.5cm/s

It is clear from above that vmII>vmI.

Therefore, the wave with faster maximum speed of a point in the medium is y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t]_ and that speed is 13.5cm/s_.

(d)

To determine

The wave which is moving in x-direction.

(d)

Expert Solution
Check Mark

Answer to Problem 87P

The wave which is moving in x-direction is y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t]_.

Explanation of Solution

The given wave equations are,

y(x,t)=(1.50cm)sin[(4.00cm1)x+(6.00s1)t]

y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t]

Write the general expression for the travelling waves.

y(x,t)=Asin[kx±ωt]

In the general wave equation, kxωt indicates a wave traveling in the positive x-direction and kx+ωt indicates a wave traveling in the negative x-direction.

From the given wave equations, y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t] has the form of the wave travelling in positive x-direction.

Conclusion:

Therefore, the wave which is moving in x-direction is y(x,t)=(4.50cm)sin[(3.00cm1)x(3.00s1)t]_.

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Chapter 11 Solutions

Physics

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