Question 1

Consider a pipe of length L that is open at both ends.What are the wavelengths of the three lowest-pitch tones produced by this pipe?

Accepted Answer

The wavelengths of the three lowest-pitch tones produced by an open pipe are given by:
λn = 2L/n
where n is the mode or harmonic number (1 for the fundamental, 2 for the first overtone, 3 for the second overtone, and so on). Substituting n = 1, 2, and 3, we get:
λ1 = 2L/1 = 2L
λ2 = 2L/2 = L
λ3 = 2L/3
Therefore, the correct choice is C, which gives the wavelengths as 2L, L, and 2L/3.

Question 2

A 2.0-m string is fixed at both ends and tightened until the wave speed is  What is the frequency of the standing wave shown in the figure?

Accepted Answer

A

Question 3

A wave pulse traveling to the right along a thin cord reaches a discontinuity where the rope becomes thinner and lighter.What is the orientation of the reflected and transmitted pulses?

Accepted Answer

A

Question 4

A platinum wire that is 1.20 m long has a radius of 0.500 mm and is fixed at both ends.In its third harmonic it vibrates at 512 Hz.The density of platinum is 21.4 × 10³ kg/m³.What is the tension in the wire?

Accepted Answer

The formula for the frequency of the nth harmonic of a fixed string is:

f(n) = n(v/2L)

where v is the speed of sound in the string, L is the length of the string and n is an integer.

For the third harmonic, n=3, so:

f(3) = 3(v/2L)

v = f(3) x 2L/3

The speed of sound in a string is given by the formula:

v = sqrt(T/μ)

where T is the tension in the string and μ is the linear density (mass per unit length) of the string. We can rearrange this formula to give T:

T = μv^2

Substituting the expression for v into this formula, we get:

T = μ(f(3)x2L/3)^2

The linear density of the wire, μ, is given by:

μ = ρV

where ρ is the density of the material and V is the volume of the wire, which is given by:

V = πr^2L

Substituting in the values given, we get:

V = π(0.0005 m)^2 x 1.20 m = 2.83x10^-7 m^3

μ = (21.4 x 10^3 kg/m^3)(2.83 x 10^-7 m^3) = 6.06 x 10^-4 kg

Substituting these values into the expression for T, we get:

T = (6.06 x 10^-4 kg)(512 Hz x 2(1.20 m)/3)^2 = 2820 N

Therefore, the tension in the wire is 2.82 kN, which is choice C.

Question 5

The lowest-pitch tone to resonate in a pipe of length L that is closed at one end and open at the other end is 200 Hz.Which one of the following frequencies will NOT resonate in the same pipe?

Accepted Answer

The answer of The lowest-pitch tone to resonate in a...

Question 6

A thin 2.00-m string of mass 50.0 g is fixed at both ends and under a tension of 70.0 N.If it is set into small-amplitude oscillation,what is the frequency of the first harmonic mode?

Accepted Answer

The answer of A thin 2.00-m string of mass 50.0...

Question 7

In a resonating pipe that is open at one end and closed at the other end,there

Accepted Answer

When a pipe is open at one end and closed at the other end, a displacement node is formed at the closed end because the air molecules cannot vibrate or move at the closed end due to its fixed boundary condition. At the open end, there is a displacement antinode because the air molecules can freely move and vibrate in this region. Therefore, the correct choice is D, which says that there is a displacement node at the closed end and a displacement antinode at the open end.

Question 8

Two pure tones are sounded together and a particular beat frequency is heard.What happens to the beat frequency if the frequency of one of the tones is increased?

Accepted Answer

The answer of Two pure tones are sounded together and...

Question 9

The lowest-pitch tone to resonate in a pipe of length L that is open at both ends is 200 Hz.Which one of the following frequencies will NOT resonate in the same pipe?

Accepted Answer

The resonant frequencies of an open pipe are given by the equation f = nv/2L, where n is any integer greater than 0, v is the speed of sound, and L is the length of the pipe. Plugging in the given values, we can find that the other resonant frequencies are: 400 Hz (n=2), 600 Hz (n=3), 800 Hz (n=4), and 1000 Hz (n=5). Therefore, the frequency that will NOT resonate in the same pipe is 900 Hz, as it cannot be obtained by this equation.

Question 10

A 1.0-g string that is 0.64 m long is fixed at both ends and is under tension.This string produces a 100-Hz tone when it vibrates in the third harmonic.The speed of sound in air is 344 m/s.The tension in the string,in is closest to

Accepted Answer

The speed of a wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension in the string and $\mu$ is the linear mass density of the string. For the third harmonic on a string fixed at both ends, the frequency $f$ is given by $f = \frac{3v}{2L}$, where $L$ is the length of the string. Rearranging the formula for $v$ to solve for $T$ and substituting the given values, we find $T$ is closest to 2.8 N.

Question 11

A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm,160 cm,and 96 cm but does not resonate at any wavelengths longer than these.This pipe is

Accepted Answer

The pipe resonates at wavelengths that fit an odd number of quarter wavelengths inside its length, which is characteristic of a pipe that is open at one end and closed at the other. The given wavelengths correspond to the 1st, 3rd, and 5th harmonics, respectively, of such a pipe.

Question 12

A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0.It is vibrating in its third OVERTONE,and the equation for the vertical displacement of any point on the string is y(x,t)= (1.22 cm)sin[(14.4 m^-1)x] cos[(166 rad/s)t]. (a)What are the frequency and wavelength of the fundamental mode of this string? (b)How long is the string? (c)How fast do waves travel on this string?

Accepted Answer

The answer of A thin taut string is fixed at...

Question 13

Consider the waves on a vibrating guitar string and the sound waves the guitar produces in the surrounding air.The string waves and the sound waves must have the same

Accepted Answer

The frequency of the waves on the guitar string and the sound waves produced in the air must be the same because the frequency of vibration of the string determines the frequency of the sound waves produced.

Question 14

A wave pulse traveling to the right along a thin cord reaches a discontinuity where the rope becomes thicker and heavier.What is the orientation of the reflected and transmitted pulses?

Accepted Answer

When a wave pulse reaches a discontinuity where the rope becomes thicker and heavier, some of the wave energy is reflected back and some is transmitted forward. When the wave energy is reflected back, the reflected pulse will be inverted because of the change in the density of the medium. The transmitted pulse will be right side up because the pulse is moving into a region of lower density where the speed of the wave is higher. Therefore, the correct choice is C, which states that the reflected pulse returns inverted while the transmitted pulse is right side up.

Question 15

A guitar string is fixed at both ends.If you tighten it to increase its tension

Accepted Answer

When you tighten a guitar string, you increase its tension. This leads to an increase in the speed of sound waves traveling through the string, which in turn leads to an increase in the frequency of the string's vibrational modes. However, the wavelength of the waves is determined by the length of the string and the boundary conditions (in this case, fixed ends), and is not affected by the tension. Therefore, choice A is the correct answer.

Question 16

In a resonating pipe that is open at both ends,there

Accepted Answer

In a resonating pipe that is open at both ends, there are displacement antinodes at each end because the air at the ends can move freely, creating maximum displacement at these points.

Question 17

A tube open at one end and closed at the other end produces sound having a fundamental frequency of 350 Hz.If you now open the closed end,the fundamental frequency becomes

Accepted Answer

When a tube open at one end and closed at the other end produces a fundamental frequency of 350 Hz, the wavelength of the sound wave in the tube is four times the length of the tube. When the closed end is opened, the wavelength of the sound wave in the tube becomes twice the length of the tube. Therefore, the new fundamental frequency would be half of the original frequency, which is 175 Hz.

Question 18

A standing wave is oscillating at 690 Hz on a string,as shown in the figure.What is the speed of traveling waves on this string?

Accepted Answer

The answer of A standing wave is oscillating at 690...

Question 19

A 2.00-m long piano wire with a mass per unit length of 12.0 g/m is under a tension of 8.00 kN.What is the frequency of the fundamental mode of vibration of this wire?

Accepted Answer

The answer of A 2.00-m long piano wire with a...

Question 20

A guitar string 0.650 m long has a tension of 61.0 N and a mass per unit length of 3.00 g/m.
(a)What is the speed of waves on the string when it is plucked?
(b)What is the string's fundamental frequency of vibration when plucked?

Accepted Answer

The answer of A guitar string 0.650 m long has...

Quiz 17: Superposition