Schaum’s Outline of College Physics

Eugene Hecht

Chapter 22

Wave Motion - all with Video Answers

Educators

Chapter Questions

Problem 1

Suppose that represents a 50 -Hz wave on a string. Take distance $y_{0}$ to be $3.0 \mathrm{~mm}$, and distance $A E$ to be $40 \mathrm{~cm}$. Find the following for the wave: its $(a)$ amplitude, $(b)$ wavelength, and $(c)$ speed.
(a) By definition, the amplitude is distance $y_{0}$ and is $3.0 \mathrm{~mm}$.
(b) The distance between adjacent crests is the wavelength, and so $\lambda=20 \mathrm{~cm}$
(c) $v=\lambda f=(0.20 \mathrm{~m})\left(50 \mathrm{~s}^{-1}\right)=10 \mathrm{~m} / \mathrm{s}$

Ma Ednelyn Lim

Ma Ednelyn Lim

Numerade Educator

Problem 2

Measurements show that the wavelength of a sound wave in a certain material is $18.0 \mathrm{~cm}$. The frequency of the wave is $1900 \mathrm{~Hz}$. What is the speed of the sound wave?
From $\lambda=v T=v / f$, which applies to all waves,
$$
u=\lambda f=(0.180 \mathrm{~m})\left(1900 \mathrm{~s}^{-1}\right)=342 \mathrm{~m} / \mathrm{s}
$$

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 3

A horizontal cord $5.00 \mathrm{~m}$ long has a mass of $1.45 \mathrm{~g}$. What must be the tension in the cord if the wavelength of a 120 -Hz wave on it is to be $60.0 \mathrm{~cm}$ ? How large a mass must be hung from its end (say, over a pulley) to give it this tension?

We know that the speed of a wave on a rope depends on both the tension and the mass per unit length. Moreover,

The tension in the cord balances the weight of the mass hung at its
end. Therefore,
$$
F_{T}=m g \quad \text { or } \quad m=\frac{F_{T}}{g}=\frac{1.50 \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=0.153 \mathrm{~kg}
$$

Ma Ednelyn Lim

Numerade Educator

Problem 4

A uniform flexible cable is $20 \mathrm{~m}$ long and has a mass of $5.0$ kg. It hangs vertically under its own weight and is vibrated (perpendicularly) from its upper end with a frequency of $7.0 \mathrm{~Hz}$.
(a) Find the speed of a transverse wave on the cable at its midpoint. (b) What are the frequency and wavelength at the midpoint?
(b) Because wave crests do not pile up along a string or cable, the number passing one point must be the same as that for any other point. Therefore, the frequency, $7.0 \mathrm{~Hz}$, is the same at all points.
To find the wavelength at the midpoint, we must use the speed we found for that point, $9.9 \mathrm{~m} / \mathrm{s}$. That gives us
$$
\lambda=\frac{v}{f}=\frac{9.9 \mathrm{~m} / \mathrm{s}}{7.0 \mathrm{~Hz}}=1.4 \mathrm{~m}
$$

Ma Ednelyn Lim

Numerade Educator

Problem 5

Suppose that depicts standing waves on a metal string under a tension of $88.2 \mathrm{~N}$. Its length is $50.0 \mathrm{~cm}$ and its mass is $0.500$ g. ( $a$ ) Compute $v$ for transverse waves on the string. (b) Determine the frequencies of its fundamental, first overtone, and second overtone.
(a) $v=\sqrt{\frac{\text { Tension }}{\sqrt{\text { Mass per t unit length }}}}=\sqrt{\frac{88.2 \mathrm{~N}}{15.00 \times 10^{-4} \mathrm{~kg} /(0.500 \mathrm{~m})}}=297 \mathrm{~m} / \mathrm{s}$
(b) We recall that the length of the segment is $\lambda / 2$ and we use $\lambda=$ $\mathrm{u} / \mathrm{f}$. For the fundamental:
$$
\lambda=1.00 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{1.00 \mathrm{~m}}=297 \mathrm{~Hz}
$$
For the first overtone:
$$
\lambda=0.500 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{0.500 \mathrm{~m}}=594 \mathrm{~Hz}
$$
For the second overtone:
$$
\lambda=0.333 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{0.333 \mathrm{~m}}=891 \mathrm{~Hz}
$$

Ma Ednelyn Lim

Numerade Educator

Problem 6

A string $2.0 \mathrm{~m}$ long is driven by a 240 -Hz vibrator at its end. The string resonates in four segments forming a standing wave pattern. What would be the speed of a transverse wave on such a string?
Let's first determine the wavelength of the wave from part (d) of Since each segment is $\lambda / 2$ long,
$$
4\left(\frac{\lambda}{2}\right)=L \quad \text { or } \quad \lambda=\frac{L}{2}=\frac{2.0 \mathrm{~m}}{2}=1.0 \mathrm{~m}
$$
Then, using $\lambda=u T=v / f$,
$$
\mathrm{u}=f \lambda=\left(240 \mathrm{~s}^{-1}\right)(1.0 \mathrm{~m})=0.24 \mathrm{~km} / \mathrm{s}
$$

Ma Ednelyn Lim

Numerade Educator

Problem 7

A banjo string $30 \mathrm{~cm}$ long oscillates in a standing-wave pattern. It resonates in its fundamental mode at a frequency of $256 \mathrm{~Hz}$. What is the tension in the string if $80 \mathrm{~cm}$ of the string have a mass of $0.75 \mathrm{~g}$ ?
First we'll find $u$ and then the tension. The string vibrates in one segment when $f=256 \mathrm{~Hz}$. Therefore, from (\mathrm{a}) \text { : }}$
and
$$
$$
\frac{0.75 \times 10^{-3} \mathrm{~kg}}{0.80 \mathrm{~m}}=9.4 \times 10^{-4} \mathrm{~kg} / \mathrm{m}
$$
Then, from, $v=\sqrt{(\text { Tension }) /(\text { Mass per unit length })}$,
$$
F_{T}=(154 \mathrm{~m} / \mathrm{s})^{2}\left(9.4 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\right)=22 \mathrm{~N}
$$
\begin{array}{l}
\frac{\lambda}{2}=L \quad \text { or } \quad \lambda=(0.30 \mathrm{~m})(2)=0.60 \mathrm{~m} \\
v=f \lambda=\left(256 \mathrm{~s}^{-1}\right)(0.60 \mathrm{~m})=154 \mathrm{~m} / \mathrm{s}
\end{array}
$$
The mass per unit length of the string is

Ma Ednelyn Lim

Numerade Educator

Problem 8

A string vibrates in five segments at a frequency of $460 \mathrm{~Hz} .(a)$ What is its fundamental frequency? (b) What frequency will cause it to vibrate in three segments?
If the string is $n$ segments long, then fromwe have $n\left(\frac{1}{2} \lambda\right)$ $=L$. But $\lambda=v / f_{n}$, so $L=n\left(\mathrm{u} / 2 f_{n}\right)$. Solving for $f_{n}$ provides
$$
f_{n}=n\left(\frac{v}{2 L}\right)
$$
We are told that $f_{5}=460 \mathrm{~Hz}$, and so
$$
460 \mathrm{~Hz}=5\left(\frac{v}{2 L}\right) \quad \text { or } \quad \frac{v}{2 L}=92.0 \mathrm{~Hz}
$$
Substituting this in the above relation gives
$$
f_{n}=(n)(92.0 \mathrm{~Hz})
$$
(a) $f_{1}=92.0 \mathrm{~Hz}$.
(b) $f_{3}=(3)(92 \mathrm{~Hz})=276 \mathrm{~Hz}$
Alternative Method
Recall that for a string held at both ends, $f_{n}=n_{f 1}$. Knowing that $f_{5}$ $=460 \mathrm{~Hz}$, it follows that $f_{1}=92.0 \mathrm{~Hz}$ and $f_{3}=276 \mathrm{~Hz}$.

Ma Ednelyn Lim

Numerade Educator

Problem 9

A string fastened at both ends resonates at $420 \mathrm{~Hz}$ and $490 \mathrm{~Hz}$ with no resonant frequencies in between. Find its fundamental resonant
frequency.
In general, $f_{n}=n f_{1}$. We are told that $f_{n}=420 \mathrm{~Hz}$ and $f_{n+1}=490 \mathrm{~Hz}$. Therefore,
$$
420 \mathrm{~Hz}=n f_{1} \text { and } 490 \mathrm{~Hz}=(n+1) f_{1}
$$
Subtract the first equation from the second to obtain $f_{1}=70.0 \mathrm{~Hz}$.

Ma Ednelyn Lim

Numerade Educator

Problem 10

A violin string resonates at its fundamental frequency of $196 \mathrm{~Hz}$. Where along the string must you place your finger so its fundamental becomes $440 \mathrm{~Hz}$ ?
For the fundamental, $L=\frac{1}{2} \lambda$. Since $\lambda=\mathrm{v} / f$, it follows that $f_{1}=$ $\mathrm{u} / 2 \mathrm{~L}$. Originally, the string of length $L_{1}$ resonated at a frequency of $196 \mathrm{~Hz}$, and therefore
$$
196 \mathrm{~Hz}=\frac{v}{2 L_{1}}
$$
with a resonance at $440 \mathrm{~Hz}$,
$$
440 \mathrm{~Hz}=\frac{v}{2 L_{2}}
$$
Eliminate $u$ from these two simultaneous equations and find
$$
\frac{L_{2}}{L_{1}}=\frac{196 \mathrm{~Hz}}{440 \mathrm{~Hz}}=0.445
$$
To obtain the desired resonance, the finger must shorten the string to $0.445$ of its original length.

Ma Ednelyn Lim

Numerade Educator

Problem 11

A 60-cm-long bar, clamped at its middle, is vibrated lengthwise by an alternating force at its end. Its fundamental resonance frequency is found to be $3.0 \mathrm{kHz}$. What is the speed of longitudinal waves in the bar?

Because its ends are free, the bar must have antinodes there. The clamp point at its center must be a node. Therefore, the
fundamental resonance is as shown in Because the distance from node to antinode is always $\frac{1}{4} \lambda$, we see that $L=2\left(\frac{1}{4} \lambda\right)$. Since $L=0.60 \mathrm{~m}$, we find $\lambda=1.20 \mathrm{~m}$.
Then, from the basic relation (p. 274) $\lambda=\mathrm{v} / f$, we have
$$
u=\lambda f=(1.20 \mathrm{~m})(3.0 \mathrm{kHz})=3.6 \mathrm{~km} / \mathrm{s}
$$

Ma Ednelyn Lim

Numerade Educator

Problem 12

Compression waves (sound waves) are sent down an air-filled tube $90 \mathrm{~cm}$ long and closed at one end. The tube resonates at several frequencies, the lowest of which is $95 \mathrm{~Hz}$. Find the speed of sound waves in air.

The tube and several of its resonance forms are shown in Recall that the distance between a node and an adjacent antinode is $\lambda / 4$. In our case, the top resonance form applies, since the segments are longest for it and its frequency is therefore lowest. For that form, $L=\lambda / 4$, so
$$
\lambda=4 L=4(0.90 \mathrm{~m})=3.6 \mathrm{~m}
$$
Using $\lambda=v T=v / f$ gives
$$
v=\lambda f=(3.6 \mathrm{~m})\left(95 \mathrm{~s}^{-1}\right)=0.34 \mathrm{~km} / \mathrm{s}
$$

Ma Ednelyn Lim

Numerade Educator

Problem 13

At what other frequencies will the tube described in underline resonate? The first few resonances are shown in We see that, at resonance,
$\begin{array}{ll}0 & \frac{1}{4} \lambda \\ A & N\end{array} \quad L=\frac{1}{4} \lambda$
(a)
$\begin{array}{llll}0 & \frac{1}{4} \lambda & 2\left(\frac{1}{4} \lambda\right) & 3\left(\frac{1}{4} \lambda\right) \\ A & N & A & N \\ L=3\left(\frac{1}{4} \lambda\right)\end{array}$
(b)
\begin{tabular}{llllll}
0 & $\frac{1}{4} \lambda$ & $2\left(\frac{1}{4} \lambda\right)$ & $3\left(\frac{1}{4} \lambda\right)$ & $4\left(\frac{1}{4} \lambda\right)$ & $5\left(\frac{1}{4} \lambda\right)$ & \\
\hline$A$ & $N$ & $A$ & $N$ & $A$ & $N$ \\
\hline
\end{tabular}$\quad L=5\left(\frac{1}{4} \lambda\right)$
$(c)$
where $n=1,3,5,7, \ldots$, is an odd integer, and $\lambda_{n}$ is the resonant wavelength. But $\lambda_{n}=\mathrm{v} / f_{n}$, and so
$$
L=n \frac{v}{4 f_{n}} \quad \text { or } \quad f_{n}=n \frac{v}{4 L}=n f_{1}
$$
where, from $\underline{\text { Problem } 22.12,} f_{1}=95 \mathrm{~Hz}$. The first few resonant frequencies are thus $95 \mathrm{~Hz}, 0.29 \mathrm{kHz}, 0.48 \mathrm{kHz}, \ldots$

Ma Ednelyn Lim

Numerade Educator

Problem 14

A metal rod $40 \mathrm{~cm}$ long is dropped, end first, onto a wooden floor and rebounds into the air. Compression waves of many frequencies are thereby set up in the bar. If the speed of compression waves in the bar is $5500 \mathrm{~m} / \mathrm{s}$, to what lowest-
frequency compression wave will the bar resonate as it rebounds?
Both ends of the bar will be free, and so antinodes will exist there. In the lowest resonance form (i.e., the one with the longest segments), only one node will exist on the bar, at its center, as illustrated in.We will then have
$$
L=2\left(\frac{\lambda}{4}\right) \quad \text { or } \quad \lambda=2 L=2(0.40 \mathrm{~m})=0.80 \mathrm{~m}
$$
Then, from $\lambda=v T=\mathrm{v} / \mathrm{f}$,
$$
f=\frac{v}{\lambda}=\frac{5500 \mathrm{~m} / \mathrm{s}}{0.80 \mathrm{~m}}=6875 \mathrm{~Hz}=6.9 \mathrm{kHz}
$$

Ma Ednelyn Lim

Numerade Educator

Problem 15

A rod $200 \mathrm{~cm}$ long is clamped $50 \mathrm{~cm}$ from one end, as shown in It is set into longitudinal vibration by an electrical driving mechanism at one end. As the frequency of the driver is slowly increased from a very low value, the rod is first found to resonate at $3 \mathrm{kHz}$. What is the speed of sound (compression waves ) in the rod?
The clamped point remains stationary, and so a node exists there. Since the ends of the rod are free, antinodes exist there. The lowest-frequency resonance occurs when the rod is vibrating in its
longest possible segments. In we show the mode of vibration that corresponds to this condition. Since a segment is the length from one node to the next, then the length from $A$ to $N$ in the figure is one-half segment. Therefore, the rod is two segments long. This resonance form satisfies our restrictions about positions of nodes and antinodes, as well as the condition that the bar vibrate in the longest segments possible. Since one segment is $\lambda / 2$ long,
$$
L=2(\lambda / 2) \text { or } \lambda=L=200 \mathrm{~cm}
$$
Then, from $\lambda=v T=\mathrm{v} / \mathrm{f}$,
$$
u=\lambda f=(2.00 \mathrm{~m})\left(3 \times 10^{3} \mathrm{~s}^{-1}\right)=6 \mathrm{~km} / \mathrm{s}
$$

Khoobchandra Agrawal

Khoobchandra Agrawal

Numerade Educator

Problem 16

(a) Determine the shortest length of pipe closed at one end that will resonate in air when driven by a sound source of frequency $160 \mathrm{~Hz}$. Take the speed of sound in air to be $340 \mathrm{~m} / \mathrm{s} .(b)$ Repeat the analysis for a pipe open at both ends.
(a) Figure $22-4(a)$ applies in this case. The shortest pipe will be $\lambda / 4$ long. Therefore,
$$
L=\frac{1}{4} \lambda=\frac{1}{4}\left(\frac{v}{f}\right)=\frac{340 \mathrm{~m} / \mathrm{s}}{4\left(160 \mathrm{~s}^{-1}\right)}=0.531 \mathrm{~m}
$$
(b) In this case the pipe will have antinodes at both ends and a node at its center. Then,
$$
L=2\left(\frac{1}{4} \lambda\right)=\frac{1}{2}\left(\frac{v}{f}\right)=\frac{340 \mathrm{~m} / \mathrm{s}}{2\left(160 \mathrm{~s}^{-1}\right)}=1.06 \mathrm{~m}
$$

Ma Ednelyn Lim

Numerade Educator

Problem 17

A pipe $90 \mathrm{~cm}$ long is open at both ends. How long must a second pipe, closed at one end, be if it is to have the same fundamental resonance frequency as the open pipe?
The two pipes and their fundamental resonances are shown in. As can be seen in the diagram,
$$
L_{o}=2\left(\frac{1}{4} \lambda\right) \quad L_{c}=\frac{1}{4} \lambda
$$
from which $L_{c}=\frac{1}{2} L_{o}=45 \mathrm{~cm}$

Ma Ednelyn Lim

Numerade Educator

Problem 18

A glass tube that is $70.0 \mathrm{~cm}$ long is open at both ends. Find the frequencies at which it will resonate when driven by sound waves that have a speed of $340 \mathrm{~m} / \mathrm{s}$.
A pipe that is open at both ends must have an antinode at each end. It will therefore resonate as in From the diagram it can be seen that the resonance wavelengths $\lambda_{n}$ are given by
$$
L=n\left(\frac{\lambda_{n}}{2}\right) \quad \text { or } \quad \lambda_{n}=\frac{2 L}{n}
$$
where $n$ is an integer. But $\lambda_{n}=v / f_{n}$, therefore
$$
f_{n}=\left(\frac{n}{2 L}\right)(v)=(n)\left(\frac{340 \mathrm{~m} / \mathrm{s}}{2 \times 0.700 \mathrm{~m}}\right)=243 n \mathrm{~Hz}
$$

Ma Ednelyn Lim

Numerade Educator

Problem 19

The average person can hear sound waves ranging in frequency from about $20 \mathrm{~Hz}$ to $20 \mathrm{kHz}$. Determine the wavelengths at these limits, taking the speed of sound to be $340 \mathrm{~m} / \mathrm{s}$.

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 20

Radio station WJR broadcasts at $760 \mathrm{kHz}$. The speed of radio waves is $3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$. What is the wavelength of WJR's waves?

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 21

Radar waves with $3.4 \mathrm{~cm}$ wavelength are sent out from a transmitter. Their speed is $3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$. What is their frequency?

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 22

A string has its tension doubled; all else kept constant, what happens to the speed of transverse waves that can be set up on the string?

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 23

A string has its total mass doubled; all else kept constant, what happens to the speed of transverse waves that can be set up on the string?

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 24

A string has both its total mass and length doubled; all else kept constant, what happens to the speed of transverse waves that can be set up on the string?

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 25

A transverse wave is set up on a taut string. Its free end is wiggled up and down at a rate of $10.0$ cycles every second. What happens to the wavelength of the waves when the oscillation rate is raised to $20.0$ cycles per second, all else kept constant?

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 26

A light cord $10.0 \mathrm{~m}$ long has a mass of $50.0 \mathrm{~g}$. It hangs vertically off the roof of a building. A man holding the bottom end of the cord pulls down on it with a force of $200 \mathrm{~N}$. He flicks the end horizontally sending a transverse pulse up the cord. Ignore the weight of the cord and determine the speed of the wave.

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 27

When driven by a 120 -Hz vibrator, a string has transverse waves of $31 \mathrm{~cm}$ wavelength traveling along it. $(a)$ What is the speed of the waves on the string? $(b)$ If the tension in the string is $1.20 \mathrm{~N}$, what is the mass of $50 \mathrm{~cm}$ of the string?

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 28

The wave shown in is being sent out by a 60-cycle/s vibrator. Find the following for the wave: $(a)$ amplitude, (b) frequency, ( $c$ ) wavelength, $(d)$ speed, (e) period.

Ma Ednelyn Lim

Numerade Educator

Problem 29

A copper wire $2.4 \mathrm{~mm}$ in diameter is $3.0 \mathrm{~m}$ long and is used to suspend a 2.0-kg mass from a beam. If a transverse disturbance is sent along the wire by striking it lightly with a pencil, how fast will the disturbance travel? The density of copper is $8920 \mathrm{~kg} / \mathrm{m}^{3}$.

Ma Ednelyn Lim

Numerade Educator

Problem 30

An explosion under the ocean creates a compression wave. Given that the bulk modulus for seawater is $2.1 \mathrm{GPa}$, how fast does the wave travel? [Hint: See Table 12-1.]

Dominique Jan Tan

Dominique Jan Tan

Numerade Educator

Problem 31

Show that the units are correct in

Ma Ednelyn Lim

Numerade Educator

Problem 32

Someone bangs on the end of a long gold rod with a hammer, creating a compression wave. How fast does it travel?

Ma Ednelyn Lim

Numerade Educator

Problem 33

Show that the fundamental frequency of a taut string on a musical instrument is given by
$$
f_{1}=(1 / 2 L)\left[F_{T} /(m / L)\right]^{1 / 2}
$$

Ma Ednelyn Lim

Numerade Educator

Problem 34

A string $180-\mathrm{cm}$ -long resonates in a standing wave that has three segments when driven by a 270 - $\mathrm{Hz}$ vibrator. What is the speed of
the waves on the string?

Ma Ednelyn Lim

Numerade Educator

Problem 35

A string resonates in three segments at a frequency of $165 \mathrm{~Hz}$. What frequency must be used if it is to resonate in four segments?

Ma Ednelyn Lim

Numerade Educator

Problem 36

A flexible cable, $30 \mathrm{~m}$ long and weighing $70 \mathrm{~N}$, is stretched by a force of $2.0 \mathrm{kN}$. If the cable is struck sideways at one end, how long will it take the transverse wave to travel to the other end and return?

Ma Ednelyn Lim

Numerade Educator

Problem 37

A wire under tension vibrates with a fundamental frequency of 256 Hz. What would be the fundamental frequency if the wire were half as long, twice as thick, and under one-fourth the tension?

Ma Ednelyn Lim

Numerade Educator

Problem 38

Steel and silver wires of the same diameter and same length are stretched with equal tension. Their densities are $7.80 \mathrm{~g} / \mathrm{cm}^{3}$ and $10.6 \mathrm{~g} / \mathrm{cm}^{3}$, respectively. What is the fundamental frequency of the silver wire if that of the steel is $200 \mathrm{~Hz}$ ?

Ma Ednelyn Lim

Numerade Educator

Problem 39

A string has a mass of $3.0 \mathrm{~g}$ and a length of $60 \mathrm{~cm}$. What must be the tension so that when vibrating transversely its first overtone has frequency $200 \mathrm{~Hz}$ ?

Ma Ednelyn Lim

Numerade Educator

Problem 40

(a) At what point should a stretched string be plucked to make its fundamental tone most prominent? At what point should it be plucked and then at what point touched $(b)$ to make its first overtone most prominent and $(c)$ to make its second overtone most nrominent?

Ma Ednelyn Lim

Numerade Educator

Problem 41

What must be the length of an iron rod that has the fundamental frequency $320 \mathrm{~Hz}$ when clamped at its center? Assume longitudinal vibration at a speed of $5.00 \mathrm{~km} / \mathrm{s}$.

Ma Ednelyn Lim

Numerade Educator

Problem 42

A rod $120 \mathrm{~cm}$ long is clamped at the center and is stroked in such a way as to give its first overtone. Make a drawing showing the location of the nodes and antinodes, and determine at what other points the rod might be clamped and still emit the same tone.

Mahendra K

Numerade Educator

Problem 43

A metal bar $6.0 \mathrm{~m}$ long, clamped at its center and vibrating longitudinally in such a manner that it gives its first overtone, vibrates in unison with a tuning fork marked 1200 vibration/s. Compute the speed of sound in the metal.

Ma Ednelyn Lim

Numerade Educator

Problem 44

Determine the length of the shortest air column in a cylindrical jar that will strongly reinforce the sound of a tuning fork having a vibration rate of $512 \mathrm{~Hz}$. Use $v=340 \mathrm{~m} / \mathrm{s}$ for the speed of sound in air.

Ma Ednelyn Lim

Numerade Educator

Problem 45

A long, narrow pipe closed at one end does not resonate to a tuning fork having a frequency of $300 \mathrm{~Hz}$ until the length of the air column reaches $28 \mathrm{~cm} .(a)$ What is the speed of sound in air at the existing room temperature? (b) What is the next length of column that will resonate to the fork?

Ma Ednelyn Lim

Numerade Educator

Problem 46

An organ pipe closed at one end is $61.0 \mathrm{~cm}$ long. What are the frequencies of the first three overtones if $u$ for sound is $342 \mathrm{~m} / \mathrm{s}$ ?

Ma Ednelyn Lim

Numerade Educator