Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 11, Problem 83P

(a)

To determine

Identify the direction of propagation of wave.

(a)

Expert Solution
Check Mark

Answer to Problem 83P

Wave propagates in left direction.

Explanation of Solution

The equation of wave is y(x,t)=(7.00cm)cos((6.00πrad/cm)x+(20.0rad/s)t).

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)=(7.00cm)cos((6.00πrad/cm)x+(20.0rad/s)t), argument is positive.

Therefore, the wave propagates in left direction.

(b)

To determine

Identify the maximum distance up to which the particles in the medium will move from the equilibrium position.

(b)

Expert Solution
Check Mark

Answer to Problem 83P

Particles can move by 7.00cm from the equilibrium position.

Explanation of Solution

The equation of wave is y(x,t)=(7.00cm)cos((6.00πrad/cm)x+(20.0rad/s)t).

Write the standard form of equation for a wave moving along –ve x-axis.

y(x,t)=Acos(kx+ωt)

Here, y(x,t) denotes the displacement of particle at any instant t, A is the amplitude, ω is the angular frequency, and k is the wave number.

The maximum distance up to which the particles in the medium will move from the equilibrium position is called the amplitude of a wave.

Conclusion:

Compare y(x,t)=(7.00cm)cos((6.00πrad/cm)x+(20.0πrad/s)t) with the standard form.

Write the value of A.

A=7.00cm

Therefore, the particles can move by 7.00cm from the equilibrium position.

(c)

To determine

Frequency of wave.

(c)

Expert Solution
Check Mark

Answer to Problem 83P

Frequency is 10.0Hz.

Explanation of Solution

The equation of wave is y(x,t)=(7.00cm)cos((6.00πrad/cm)x+(20.0πrad/s)t).

Compare y(x,t)=(7.00cm)cos((6.00πrad/cm)x+(20.0rad/s)t) with the standard form y(x,t)=Acos(kx+ωt).

Write the value of ω.

ω=20.0πrad/s

Write the relation between the linear frequency and ω.

f=ω2π

Here, f is the frequency of wave.

Conclusion:

Substitute 20.0πrad/s for ω in the above equation to find f.

f=20.0πrad/s2π=10πHz

Therefore, the frequency is 10.0Hz.

(d)

To determine

Wavelength of wave.

(d)

Expert Solution
Check Mark

Answer to Problem 83P

Wavelength is 0.333cm.

Explanation of Solution

The equation of wave is y(x,t)=(7.00cm)cos((6.00πrad/cm)x+(20.0πrad/s)t).

Compare y(x,t)=(7.00cm)cos((6.00πrad/cm)x+(20.0rad/s)t) with the standard form y(x,t)=Acos(kx+ωt).

Write the value of k.

k=6.00πrad/cm

Write the relation between wavelength and k.

λ=2πk

Here, λ is the wavelength of wave.

Conclusion:

Substitute 6.00πrad/cm for k in the above equation to find λ.

λ=2π6.00πrad/cm=0.333cm

Therefore, the wavelength is 0.333cm.

(e)

To determine

Speed of wave.

(e)

Expert Solution
Check Mark

Answer to Problem 83P

Speed is 3.33cm/s.

Explanation of Solution

The equation of wave is y(x,t)=(7.00cm)cos((6.00πrad/cm)x+(20.0πrad/s)t).

Write the relation between the speed of wave and k.

v=ωk

Here, v is the speed of wave.

Conclusion:

Substitute 20.0πrad/s for ω and 6.00πrad/cm for k in the above equation to find v.

v=20.0πrad/s6.00πrad/cm=3.33cm/s

Therefore, the speed is 3.33cm/s.

(f)

To determine

Explain the motion at y=7.00cm and x=0 when t=0.

(f)

Expert Solution
Check Mark

Explanation of Solution

The equation of wave is y(x,t)=(7.00cm)cos((6.00πrad/cm)x+(20.0πrad/s)t).

In the given equation of wave, y(x,t) denotes the displacement of particle and the variable along with k inside the cosine function denotes the direction of propagation of wave. Cosine function shows that the particle is executing sinusoidal motion. Displacement of particle y(x,t) means the particle is moving along the y-axis.

On substituting x=0 and t=0 in the given equation, will results in the maximum value of cosine function means that the particle will be at the maximum amplitude at the moment. Since the given form equation is consistent with stand form of equation, it can be say the particle is vibrating about y=0.

Therefore, the particle is under sinusoidal oscillation with amplitude 7.00cm about y=0 along y-axis.

(g)

To determine

Check whether the wave is transverse of longitudinal in nature.

(g)

Expert Solution
Check Mark

Explanation of Solution

Longitudinal wave is one which the displacement of particles of medium and the direction of wave are in same direction. For transverse wave, displacement of particles of medium and the direction of wave will be perpendicular to each other.

In the give wave, it is found that particle is vibrating along y-axis and the wave propagates along x-axis.

Therefore, the given wave is a transverse wave.

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Chapter 11 Solutions

Physics

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