Question #24

Practice Question

Concept Applied

$v = \omega/k = (8\pi)/(4\pi) = 2\,\text{m/s}$....

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Question #25

Practice Question

A.2 m/s
B.3 m/s
C.4 m/s
D.5 m/s

Concept Applied

The wave speed is $v = \frac{\omega}{k}$ with $\omega = 2\pi\,$rad/s and $k = \pi\,$rad/m, giving $v = 2\,$m/s....

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Question #26

Practice Question

Concept Applied

From $k = 4\pi$, $\omega = 8\pi$, $v = \omega/k = 2\,\text{m/s}$....

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Question #27

Practice Question

A.0.5 m/s
B.1 m/s
C.2 m/s
D.4 m/s

Concept Applied

The general form of a wave is $ y = A \sin(\omega t - kx) $. Comparing, $ \omega = 4\pi $, $ k = 2\pi $. Wave speed $ v = \frac{\omega}{k} = \frac{4\p...

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Question #28

Practice Question

A.$ \frac{4L}{3} $
B.$ \frac{2L}{3} $
C.$ \frac{L}{3} $
D.$ \frac{3L}{2} $

Concept Applied

For a string fixed at both ends, the wavelength for the $ n^{\text{th}} $ harmonic is $ \lambda_n = \frac{2L}{n} $. For the third harmonic ($ n=3 $), ...

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Question #29

Practice Question

Concept Applied

$v \propto \sqrt{T}$, $T_1 = 273$, $T_2 = 373$, so $v_2 = 332 \times \sqrt{373/273} \approx 387$ m/s....

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Question #30

Practice Question

A.$4\pi$
B.
.$8\pi$
D.$\pi$

Concept Applied

The wave number is given by $k = 2\pi/\lambda$. Substituting $\lambda = 0.5\ \text{m}$ gives $k = 4\pi\ \text{rad\,m}^{-1}$....

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Question #31

Practice Question

A.Increasing the temperature $T$
B.Increasing the molar mass $M$
C.Decreasing the adiabatic index $\gamma$
D.Decreasing the temperature $T$

Concept Applied

From $v = \sqrt{\gamma R T/M}$, $v$ is directly proportional to $\sqrt{T}$ and $\sqrt{\gamma}$, and inversely proportional to $\sqrt{M}$. Hence raisin...

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Question #32

Practice Question

A.252 Hz
B.254 Hz
C.258 Hz
D.260 Hz

Concept Applied

Beat frequency = |f1 - f2|. So, 256 - f2 = 4 → f2 = 252 Hz....

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Question #33

Practice Question

A.$\frac{3v}{2L}$
B.$\frac{v}{2L}$
C.$\frac{3v}{4L}$
D.$\frac{2v}{3L}$

Concept Applied

For a string fixed at both ends, the wavelength of the $n$-th harmonic is $\lambda_n = \frac{2L}{n}$. For third harmonic ($n=3$), $\lambda = \frac{2L}...

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Question #34

Practice Question

A.$\frac{L}{2}$
B.$L$
C.$\frac{3L}{2}$
D.
...

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Question #35

Practice Question

A.$660\,\text{Hz}$
B.$650\,\text{Hz}$
C.$550\,\text{Hz}$
D.$600\,\text{Hz}$

Concept Applied

Use Doppler formula: $f' = f \frac{v}{v - v_s} = 600 \times \frac{330}{300} = 660\,\text{Hz}$....

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Question #36

Practice Question

Concept Applied

Beat frequency is absolute difference: $|260 - 256| = 4\,\text{Hz}$....

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Question #37

Practice Question

Concept Applied

Constructive interference occurs when phase difference is 0, 2π, 4π,... radians....

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Question #38

Practice Question

A.141.7\ \text{Hz}
B.283.3\ \text{Hz}
C.340\ \text{Hz}
D.170\ \text{Hz}

Concept Applied

For a pipe open at both ends, the fundamental wavelength is $\lambda = 2L$. Hence $f = v/\lambda = v/(2L) = 340/(2\times1.2) \approx 141.7\ \text{Hz}$...

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Question #39

Practice Question

A.$\sqrt{2}$
B.2
C.$\frac{1}{\sqrt{2}}$
D.1

Concept Applied

Since $v \propto \sqrt{T}$ for a given gas, the ratio of the new speed to the original speed is $\sqrt{600/300}=\sqrt{2}$. Hence the speed of sound in...

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Question #40

Practice Question

A.$|f_1-f_2|$
B.$f_1+f_2$
C.
.
...

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Question #41

Practice Question

Concept Applied

$v = \omega/k = (8\pi)/(4\pi) = 2\,\text{m/s}$....

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Question #42

Practice Question

A.2 m/s
B.3 m/s
C.4 m/s
D.5 m/s

Concept Applied

The wave speed is $v = \frac{\omega}{k}$ with $\omega = 2\pi\,$rad/s and $k = \pi\,$rad/m, giving $v = 2\,$m/s....

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Question #43

Practice Question

Concept Applied

From $k = 4\pi$, $\omega = 8\pi$, $v = \omega/k = 2\,\text{m/s}$....

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Question #44

Practice Question

A.0.5 m/s
B.1 m/s
C.2 m/s
D.4 m/s

Concept Applied

The general form of a wave is $ y = A \sin(\omega t - kx) $. Comparing, $ \omega = 4\pi $, $ k = 2\pi $. Wave speed $ v = \frac{\omega}{k} = \frac{4\p...

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Question #45

Practice Question

A.$ \frac{4L}{3} $
B.$ \frac{2L}{3} $
C.$ \frac{L}{3} $
D.$ \frac{3L}{2} $

Concept Applied

For a string fixed at both ends, the wavelength for the $ n^{\text{th}} $ harmonic is $ \lambda_n = \frac{2L}{n} $. For the third harmonic ($ n=3 $), ...

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Question #46

Practice Question

Concept Applied

$v \propto \sqrt{T}$, $T_1 = 273$, $T_2 = 373$, so $v_2 = 332 \times \sqrt{373/273} \approx 387$ m/s....

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Question #47

Practice Question

A.$4\pi$
B.
.$8\pi$
D.$\pi$

Concept Applied

The wave number is given by $k = 2\pi/\lambda$. Substituting $\lambda = 0.5\ \text{m}$ gives $k = 4\pi\ \text{rad\,m}^{-1}$....

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Question #48

Practice Question

A.Increasing the temperature $T$
B.Increasing the molar mass $M$
C.Decreasing the adiabatic index $\gamma$
D.Decreasing the temperature $T$

Concept Applied

From $v = \sqrt{\gamma R T/M}$, $v$ is directly proportional to $\sqrt{T}$ and $\sqrt{\gamma}$, and inversely proportional to $\sqrt{M}$. Hence raisin...

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Question #49

Practice Question

A.252 Hz
B.254 Hz
C.258 Hz
D.260 Hz

Concept Applied

Beat frequency = |f1 - f2|. So, 256 - f2 = 4 → f2 = 252 Hz....

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Question #50

Practice Question

A.$\frac{3v}{2L}$
B.$\frac{v}{2L}$
C.$\frac{3v}{4L}$
D.$\frac{2v}{3L}$

Concept Applied

For a string fixed at both ends, the wavelength of the $n$-th harmonic is $\lambda_n = \frac{2L}{n}$. For third harmonic ($n=3$), $\lambda = \frac{2L}...

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