Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 11, Problem 38P

(a)

To determine

Identify the direction of propagation of wave.

(a)

Expert Solution
Check Mark

Answer to Problem 38P

Wave propagates in left direction.

Explanation of Solution

Speed of transverse wave is 10.0m/s and the equation of wave is y(x,t)=Acos(ωt+kx).

The direction of wave is determined by whether the argument of cosine function is positive or not. If it is positive, means that the wave is propagating in left direction and negative value indicates the propagation in right direction. For the wave y(x,t)=Acos(ωt+kx), argument is positive.

Therefore, the wave propagates in left direction.

(b)

To determine

The numerical values of A,ω and k.

(b)

Expert Solution
Check Mark

Answer to Problem 38P

Ais2.0mm, ωis1600rad/sandkis160rad/s.

Explanation of Solution

Speed of transverse wave is 10.0m/s and the equation of wave is y(x,t)=Acos(ωt+kx).

Amplitude is the maximum displacement from the mean position. It can be found directly from the diagram.

Write the value of amplitude.

A=2.0mm

Write the equation to find the angular frequency.

ω=2πvλ

Here, ω is the angular frequency, v is the speed f wave, and λ is the wavelength.

Wavelength is the distance from beginning of a crest to the end of adjacent trough. From the graph it is found that the wavelength is 4.0cm.

Write the equation to find the wave number.

k=2πλ (I)

Here, k is the wave number.

Conclusion:

Substitute 3.14 for π, 10.0m/s for v, and 4.0cm for λ in the equation for ω.

ω=2(3.14)(10.0m/s)4.0cm(1m100cm)=1600rad/s

Substitute 3.14 for π and 4.0cm for λ in the equation for ω.

k=2(3.14)4.0cm(1m100cm)=160rad/m

Therefore, Ais2.0mm, ωis1600rad/sandkis160rad/s.

(c)

To determine

Identify the instants (three smallest non negative cases) at which the snap of wave is taken.

(c)

Expert Solution
Check Mark

Answer to Problem 38P

The snap is shot at 1.0ms,5.0ms,9.0ms.

Explanation of Solution

Speed of transverse wave is 10.0m/s and the equation of wave is y(x,t)=Acos(ωt+kx).

Time of snap shot can be identified from the instants at which maximum displacement occurs.

Write the value of cosine function at the time of maximum displacement.

cos(ωt+kx)=1

Write the possible values for (ωt+kx) to satisfy the above condition.

ωt+kx=2nπ

Here, n is any integer.

Rewrite the above relation in terms of t.

t=(2nπkx)ω . (II)

It is seen that there is a peak at 3.0cm when the snapshot is obtained.

Rewrite equation (I) by multiplying with x on both sides.

kx=2πλx

Substitute 4.0cm for λ and 3.0cm for x in the above equation.

kx=2π4.0cm(3.0cm)=3π2

Rewrite equation (II) by substituting the above value of kx.

t=(2nπ3π2)ω

Write the relation between the time period and ω.

ω=2πT

Here, T is the time period of wave.

Rewrite the previous equation for t by substituting the above relation for ω.

t=(2nπ3π2)(2πT)=(n34)T (III)

Write the relation between T and v.

T=λv (IV)

Conclusion:

Substitute 4.0cm for λ and 10.0m/s for v in equation (IV).

T=4.0cm(1m100cm)10.0m/s=4.0×103s(1ms103s)=4.0ms

Substitute 1 for n and 4.0ms for T in equation (III) to find t.

t=(134)(4.0ms)=1.0ms

Substitute 2 for n and 4.0ms for T in equation (III) to find t.

t=(234)(4.0ms)=5.0ms

Substitute 3 for n and 4.0ms for T in equation (III) to find t.

t=(334)(4.0ms)=9.0ms

Therefore, the snap is shot at 1.0ms,5.0ms,9.0ms.

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Chapter 11 Solutions

Physics

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