Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
5th Edition
ISBN: 9781305586871
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 14, Problem 28P

(a)

To determine

The number of loops in the pattern

(a)

Expert Solution
Check Mark

Answer to Problem 28P

The pattern exhibits a total of 3 loops.

Explanation of Solution

Write the expression for the wave function of a standing wave.

    y=2Asinkxcosωt        (I)

Here, A is the amplitude of the standing wave, k is the wave number and ω is the angular frequency of the standing wave.

Write the given equation for the wave function of standing waves

    y=0.002sin(πx)cos(100πt)        (II)

Compare equation (I) and equation (II) to find the value of wave number.

    k=πm-1        (III)

Write the equation for the wave number.

    k=2πλ        (IV)

Here, λ is the wavelength of the standing wave. Compare equation (III) and equation (IV) to find the wavelength of the standing wave.

    2πλ=πm-1λ=2.00m

Compare equation (I) and equation (II) to find the value of the angular frequency

    ω=100πs-1        (V)

Write the expression for the angular frequency.

    ω=2πf        (VI)

Here, f is the frequency of the standing wave. Compare the equation (V) and (VI) to find the frequency of the standing wave.

    2πf=100πs-1f=100πs-12π=50Hz

Conclusion:

The distance between two adjacent nodes is half the wavelength of the standing wave. Write the equation for the distance between two adjacent nodes.

    dNN=λ2

Substitute 2.00m for λ

    dNN=2.00m2=1.00m

The number of loops is the ration between the length of the wire and the distance between the two adjacent nodes. Write the equation for the number of loops.

    n=LdNN

Here, n is the number of loops, L is the length of the wire and dNN is the distance between the two adjacent nodes. Substitute 3.00m for L and 1.00m for dNN.

    n=3.00m1.00m=3loops

Therefore, the pattern exhibits a total of 3 loops.

(b)

To determine

The fundamental frequency of vibration

(b)

Expert Solution
Check Mark

Answer to Problem 28P

The fundamental frequency of vibration of the wire is 16.7Hz.

Explanation of Solution

Refer SUB-PART (a)

Conclusion:

Write the equation for the speed of the wave.

    v=ωk

Here, ω is the angular frequency of the standing wave and k is the wave number.

Substitute 100πs-1 for ω and πm-1 for k.

    v=100πs-1πm-1=100m/s        (VII)

The simplest standing wave has the distance of the wire as the distance between the adjacent nodes as they have only two nodes, one at each end.

    dNN=3.00m        (VIII)

Also, the distance between the nodes at this case is half the wavelength of the simplest standing wave.

    dNN=λa2        (IX)

Here, λa is the wavelength of the simplest standing wave. Compare equation (VII) and equation (VIII).

    λa2=3.00m=6.00m

Write the expression for the frequency of the simplest standing wave in this case.

    fa=vλa

Substitute 100m/s for v and 6.00m for λa.

    fa=100m/s6.00m=16.7Hz

(c)

To determine

The number of loops in the new pattern

(c)

Expert Solution
Check Mark

Answer to Problem 28P

The new pattern has 1 loop.

Explanation of Solution

Write the equation for the velocity in terms of the tension on the string.

    v=Tμ        (X)

Here, T is the tension on the wire and μ is the mass per unit length of the wire.

Write the equation for the tension on the string when the tension is increased by 9 times.

    T0=9T        (XI)

Here, T0 is the new tension.

Write the new equation for the speed of the wave when the tension is increased.

    v0=T0μ        (XII)

Conclusion:

Substitute equation (XI) in equation (XII).

    v0=9Tμ=3Tμ=3v

Substitute 100m/s for v from equation (VII).

    v0=3(100m/s)=300m/s

Write the equation for the wavelength of the wave when the tension on the string is increased.

    λ0=v0f

Here, v0 is the speed of the wave after the tension is increased and f is the frequency of the wave. Substitute 300m/s for v0 and 50Hz for f.

    λ0=300m/s50Hz=6.00m

The distance between the nodes at this case is half the wavelength of the standing wave.

    dNN=λ02

Substitute 6.00m for λ0.

    dNN=6.00m2=3.00m

Therefore, the distance between two adjacent nodes when the tension is increased is 3.00m. This shows that the distance between two adjacent nodes is same as the length of the wire. Hence, there will be only one loop after the tension is increased by 9 times.

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Chapter 14 Solutions

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)

Ch. 14 - Prob. 5OQCh. 14 - Prob. 6OQCh. 14 - Prob. 7OQCh. 14 - Prob. 8OQCh. 14 - Prob. 9OQCh. 14 - Prob. 10OQCh. 14 - A standing wave having three nodes is set up in a...Ch. 14 - Prob. 1CQCh. 14 - Prob. 2CQCh. 14 - Prob. 3CQCh. 14 - Prob. 4CQCh. 14 - What limits the amplitude of motion of a real...Ch. 14 - Prob. 6CQCh. 14 - Prob. 7CQCh. 14 - Prob. 8CQCh. 14 - Prob. 1PCh. 14 - Prob. 2PCh. 14 - Prob. 3PCh. 14 - Prob. 4PCh. 14 - Prob. 5PCh. 14 - Prob. 6PCh. 14 - Prob. 7PCh. 14 - Prob. 8PCh. 14 - Prob. 9PCh. 14 - Prob. 10PCh. 14 - Prob. 11PCh. 14 - Prob. 12PCh. 14 - Prob. 13PCh. 14 - Prob. 14PCh. 14 - Prob. 15PCh. 14 - Prob. 16PCh. 14 - Prob. 17PCh. 14 - Prob. 18PCh. 14 - Prob. 19PCh. 14 - Prob. 20PCh. 14 - A string with a mass m = 8.00 g and a length L =...Ch. 14 - Prob. 22PCh. 14 - Prob. 23PCh. 14 - Prob. 24PCh. 14 - Prob. 25PCh. 14 - Review. A sphere of mass M is supported by a...Ch. 14 - Prob. 27PCh. 14 - Prob. 28PCh. 14 - Prob. 29PCh. 14 - Prob. 30PCh. 14 - Prob. 31PCh. 14 - The overall length of a piccolo is 32.0 cm. The...Ch. 14 - Prob. 33PCh. 14 - Prob. 34PCh. 14 - Two adjacent natural frequencies of an organ pipe...Ch. 14 - Do not stick anything into your ear! Estimate the...Ch. 14 - Prob. 37PCh. 14 - As shown in Figure P14.37, water is pumped into a...Ch. 14 - Prob. 39PCh. 14 - Prob. 40PCh. 14 - Prob. 41PCh. 14 - Why is the following situation impossible? A...Ch. 14 - 23. An air column in a glass tube is open at one...Ch. 14 - Prob. 44PCh. 14 - Prob. 45PCh. 14 - Prob. 46PCh. 14 - Prob. 47PCh. 14 - Prob. 48PCh. 14 - Some studies suggest that the upper frequency...Ch. 14 - Prob. 50PCh. 14 - An earthquake can produce a seiche in a lake in...Ch. 14 - Prob. 52PCh. 14 - Prob. 53PCh. 14 - Prob. 54PCh. 14 - Prob. 55PCh. 14 - A nylon string has mass 5.50 g and length L = 86.0...Ch. 14 - Prob. 57PCh. 14 - Prob. 58PCh. 14 - Prob. 59PCh. 14 - Review. For the arrangement shown in Figure...Ch. 14 - Prob. 61PCh. 14 - Prob. 62PCh. 14 - Prob. 63PCh. 14 - Prob. 64PCh. 14 - Prob. 65PCh. 14 - Prob. 66PCh. 14 - Prob. 67PCh. 14 - Review. Consider the apparatus shown in Figure...Ch. 14 - Prob. 69P
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