Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781337026345
Author: Katz
Publisher: Cengage
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Chapter 18, Problem 16PQ

As in Figure P18.16, a simple harmonic oscillator is attached to a rope of linear mass density 5.4 × 10−2 kg/m, creating a standing transverse wave. There is a 3.6-kg block hanging from the other end of the rope over a pulley. The oscillator has an angular frequency of 43.2 rad/s and an amplitude of 24.6 cm. a. What is the distance between adjacent nodes? b. If the angular frequency of the oscillator doubles, what happens to the distance between adjacent nodes? c. If the mass of the block is doubled instead, what happens to the distance between adjacent nodes? d. If the amplitude of the oscillator is doubled, what happens to the distance between adjacent nodes?

Chapter 18, Problem 16PQ, As in Figure P18.16, a simple harmonic oscillator is attached to a rope of linear mass density 5.4

FIGURE P18.16

(a)

Expert Solution
Check Mark
To determine

The distance between the adjacent nodes.

Answer to Problem 16PQ

The distance between the adjacent nodes is 1.9m_.

Explanation of Solution

Given that the linear mass density of the rope is 5.4×102kg/m, the mass of the block is 3.6kg, frequency of the oscillator is 43.2rad/s and the amplitude of the harmonic waves is 24.6cm.

Write the expression for the wavelength of the wave.

λ=vf                                                                                                                         (I)

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

Write the expression for the speed of the wave.

v=FTμ (II)

Here, FT is the force of tension on rope, μ is the linear mass density.

Write the equation to find the frequency of the wave.

f=ω2π (III)

Here, ω is the angular frequency of the wave.

Use equation (III) and (II) in (I).

λ=1fFTμ=2πωFTμ (IV)

Write the expression for the tension force on the rope (it is equal to the weight of the block hanged).

  FT=mg

Here, g is the acceleration due to gravity.

Use above expression in equation (IV).

λ=2πωmgμ (V)

The distance between the adjacent nodes is equal to half the wavelength of the wave.

  dnodes=12λ

Rewrite above equation using equation (V).

dnodes=12(2πωmgμ)                                                                                               (VI)

Conclusion:

Substitute 43.2rad/s for ω, 3.6kg for m, 9.81m/s2 for g, and 5.4×102kg/m for μ in equation (VI) to get dnodes.

  λ=12(2π43.2rad/s(3.6kg)(9.81m/s2)5.4×102kg/m)=1.9m

Therefore, the distance between the adjacent nodes is 1.9m_.

(b)

Expert Solution
Check Mark
To determine

The distance between the adjacent nodes when the angular frequency is doubled.

Answer to Problem 16PQ

The distance between the adjacent nodes when the angular frequency is doubled is 0.93m_.

Explanation of Solution

Equation (VI) gives the expression for the distance between the nodes.

  dnodes=12(2πωmgμ)

From the above equation it is clear that wavelength is inversely proportional to angular frequency.

The angular frequency is doubled. Replace ω by 2ω in equation (VI) to find the new distance dnodes.

dnodes=12(2π2ωmgμ)=(π2ωmgμ) (VII)

Conclusion:

Substitute 43.2rad/s for ω, 3.6kg for m, 9.81m/s2 for g, and 5.4×102kg/m for μ in the equation (VII) to get dnodes.

  dnodes=(π2(43.2rad/s)(3.6kg)(9.81m/s2)5.4×102kg/m)=0.93m

Therefore, distance between the adjacent nodes when the angular frequency is doubled is 0.93m_.

(c)

Expert Solution
Check Mark
To determine

The distance between the adjacent nodes if the mass of the block is doubled.

Answer to Problem 16PQ

The distance between the adjacent nodes if the mass of the block is doubled is 2.6m_.

Explanation of Solution

Equation (VI) gives the expression for the distance between the nodes.

  dnodes=12(2πωmgμ)

When the mass of the block is doubled, m becomes 2m.Replace m by 2m in equation (VI) to find the new distance dnodes.

dnodes=12(2πω2mgμ)=(4πωmgμ) (VIII)

Conclusion:

Substitute 43.2rad/s for ω, 3.6kg for m, 9.81m/s2 for g, and 5.4×102kg/m for μ in the equation (VIII) to get dnodes.

  dnodes=(4π43.2rad/s(3.6kg)(9.81m/s2)5.4×104kg/m)=2.6m

Therefore, the distance between the adjacent nodes if the mass of the block is doubled is 2.6m_.

(d)

Expert Solution
Check Mark
To determine

The distance between the nodes if the amplitude of the oscillator is doubled.

Answer to Problem 16PQ

The distance between the nodes remains the same even if the amplitude of the oscillator is doubled.

Explanation of Solution

Equation (VI) gives the expression for the distance between the nodes.

  dnodes=12(2πωmgμ)

The above equation is independent of the amplitude term. Thus, even if the amplitude of the oscillator is doubled, the distance between the nodes do not change.

Conclusion:

Therefore, the distance between the nodes remains the same even if the amplitude of the oscillator is doubled.

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Chapter 18 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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