Question 1

When the standing wave pattern in a pipe is NANA,the pipe has which of the following set of properties? (N stands for node,A for antinode. )

Accepted Answer

In a pipe with a standing wave pattern of NANA, there is a node at one end and an antinode at the other end. This is only possible if the pipe is open at one end (the node) and closed at the other end (the antinode). Therefore, choice C is the best option.

Question 2

A standing wave is set up in a 200-cm string fixed at both ends.The string vibrates in 8 distinct segments when driven by a 120-Hz source.What is the wavelength?

Accepted Answer

The length of the string is 200 cm, and it vibrates in 8 segments, meaning there are 8 half-wavelengths in the string. Therefore, the wavelength is (200 cm / 8) * 2 = 50 cm.

Question 3

The lower A on a piano has a frequency of 30.0 Hz.If the tension in the 2.00-m-long string is 316 N and one-half wavelength occupies the string,what is the mass of the string?

Accepted Answer

The answer of The lower A on a piano has...

Question 4

If a guitar string has a fundamental frequency of 512 Hz,what is the frequency of its first overtone?

Accepted Answer

The first overtone of a vibrating string is twice the frequency of its fundamental frequency. Therefore, the frequency of the first overtone in this case would be 2 x 512 Hz = 1 024 Hz. However, since the question asks for the closest answer, the closest option would be D) 1 020 Hz.

Question 5

A C note (f = 256 Hz)is sounded on a piano.If the length of the piano wire is 1.8 m and its mass density is 2.2 g/m,what is the tension in the wire?

Accepted Answer

The formula for the speed of a wave on a string is v = √(T/μ), where T is the tension and μ is the mass density.
For a C note with frequency f = 256 Hz and wavelength λ = 2L, where L is the length of the string, the speed of the wave can be calculated as v = fλ = 512 m/s.
Equating this to the formula for v above and solving for T, we get T = μv^2 = 1.9E+3 N. Therefore, the correct answer is C.

Question 6

What is the lowest frequency that will resonate in an organ pipe 5.00 m in length,closed at one end? The speed of sound in air is 340 m/s.

Accepted Answer

The fundamental frequency (first harmonic) for a pipe closed at one end is given by $f = \frac{v}{4L}$, where $v$ is the speed of sound in air (340 m/s) and $L$ is the length of the pipe (5.00 m). Substituting these values gives $f = \frac{340}{4 \times 5} = \frac{340}{20} = 17.0$ Hz.

Question 7

A tuning fork is sounded above a resonating tube (one end closed),which resonates at a length of 0.40 m and again at 1.2 m.If the tube length were extended further,at what point will the tuning fork again create a resonance condition?

Accepted Answer

The answer of A tuning fork is sounded above a...

Question 8

A 2.5-m-long organ pipe is open at one end and closed at the other.Its fundamental tone has wavelength

Accepted Answer

The answer of A 2.5-m-long organ pipe is open at...

Question 9

A 1.30-m string is held fixed at both ends.When driven by a 140-Hz source,the string vibrates in 4 distinct segments.What is the natural fundamental frequency of the string?

Accepted Answer

When the string vibrates in 4 distinct segments, it is in its 4th harmonic. The fundamental frequency (1st harmonic) is a quarter of the 4th harmonic's frequency. Given the 4th harmonic is 140 Hz, the fundamental frequency is 140 Hz / 4 = 35.0 Hz.

Question 10

An astronaut sets up a pendulum on the Moon,where the acceleration due to gravity is 1/6 that of Earth.If she finds that she must push the pendulum with a frequency of 0.10 Hz to maintain a swinging motion,what is the length of the pendulum?

Accepted Answer

The frequency of a pendulum is given by $f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$, where $g$ is the acceleration due to gravity and $L$ is the length of the pendulum. On the Moon, $g = \frac{1}{6}g_{Earth}$, and given $f = 0.10$ Hz, we can rearrange the formula to solve for $L$: $L = \frac{g}{(2\pi f)^2}$. Substituting $g_{Moon} = \frac{1}{6}g_{Earth} = \frac{1}{6} \times 9.8 \, m/s^2$ and $f = 0.10$ Hz, we find $L \approx 4.1$ m.

Question 11

A child sits on a swing supported by ropes of length 2.0 m.With what frequency will she need to apply the driving force to maintain swinging?

Accepted Answer

The frequency of the driving force should be equal to the natural frequency of the swing. The natural frequency of a pendulum is given by f = 1/2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. In this case, L = 2.0 m and g = 9.8 m/s^2, so f = 1/2π√(2.0 m / 9.8 m/s^2) = 0.35 Hz.

Question 12

Quintupling the tension in a guitar string will change its natural frequency by what factor?

Accepted Answer

The natural frequency of a guitar string is directly proportional to the square root of its tension. Quintupling the tension means multiplying it by 5. Therefore, the new frequency will be the square root of 5 times the old frequency. This can be simplified as sqrt(5)*sqrt(frequency) or sqrt(5*frequency). Hence, the factor by which the natural frequency changes is sqrt(5) which is approximately 2.24. Choice C, 11ea800b_1910_5e3c_846e_657262fd77bd_TB6864_11, is the closest approximation to this value.

Question 13

A pipe is open at both ends.The distance from a node to its adjacent antinode is equal to _______.​

Accepted Answer

The distance from a node to its adjacent antinode in a pipe open at both ends (or in any standing wave) is equal to one-quarter of the wavelength (0.25λ). This is because a full wavelength (λ) consists of two nodes and two antinodes, making the distance between a node and its adjacent antinode λ/4.

Question 14

A 90-m-long high-voltage cable is suspended between two towers.The mass of the 90-m cable is 100 kg.If the tension in the cable is 25 000 N,what is the lowest frequency at which this cable can oscillate?

Accepted Answer

The lowest frequency at which the cable can oscillate, also known as the fundamental frequency, can be calculated using the formula for the fundamental frequency of a string or cable under tension, which is $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $f$ is the frequency, $L$ is the length of the cable, $T$ is the tension in the cable, and $\mu$ is the mass per unit length of the cable. Given that the length $L = 90$ m, the tension $T = 25,000$ N, and the mass of the cable is $100$ kg (thus $\mu = \frac{100}{90} = \frac{10}{9}$ kg/m), we can substitute these values into the formula to find the frequency.First, calculate the mass per unit length, $\mu = \frac{mass}{length} = \frac{100}{90} = \frac{10}{9}$ kg/m.Then, substitute the values into the formula:$$f = \frac{1}{2 \times 90} \sqrt{\frac{25,000}{10/9}}$$$$f = \frac{1}{180} \sqrt{\frac{25,000 \times 9}{10}}$$$$f = \frac{1}{180} \sqrt{22,500}$$$$f = \frac{1}{180} \times 150$$$$f = \frac{150}{180}$$$$f = \frac{5}{6}$$$$f = 0.83$$ HzTherefore, the lowest frequency at which the cable can oscillate is 0.83 Hz.

Question 15

A string with fixed ends is made to go into standing wave patterns.At a given tension,the lowest frequency for which such a pattern exists is 150 Hz.Which of the following frequencies would also cause a standing wave pattern in this case?

Accepted Answer

The answer of A string with fixed ends is made...

Question 16

A note is being played on a string on a guitar.The string is then pushed to the side while still in contact with the same fret.What happens to the frequency of the note then played compared to the original note?

Accepted Answer

Pushing the string to the side while in contact with the same fret increases the tension in the string, which increases the frequency of the vibration, resulting in a higher pitched note.

Question 17

If I quadruple the mass per unit length of guitar string,its natural frequency changes by what factor?

Accepted Answer

The answer of If I quadruple the mass per unit...

Question 18

I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency.I then stretch it to four times its length and make it vibrate in the fundamental frequency once again.The rubber band is made so that quadrupling its length quadruples the tension and reduces the mass per unit length by a factor of 4.The new frequency will be related to the old by a factor of

Accepted Answer

The answer of I stretch a rubber band and "plunk"...

Question 19

For a standing wave on a string the wavelength must equal

Accepted Answer

The wavelength of a standing wave on a string is twice the distance between adjacent nodes because one complete wavelength includes a node, an antinode, another node, and back to the first node, which is twice the node-to-node distance.

Question 20

Standing-wave vibrations are set up in a crystal wine glass with four nodes and four antinodes equally spaced around the 2.80-cm radius of the rim.If the resonant frequency of the glass is 8.50 kHz,what is the speed of the transverse waves moving around the glass?

Accepted Answer

The resonant frequency of the glass is given by:

f = (v/λ)

where v is the speed of the transverse waves around the glass and λ is the wavelength of the standing wave.

We can find the wavelength from the distance between nodes or antinodes, which is half of the wavelength:

λ = 2(2.80 cm) = 5.60 cm = 0.0560 m

Substituting the values:

8.50 kHz = (v/0.0560 m)

v = (8.50 kHz)(0.0560 m) = 0.476 m/s

Therefore, the speed of the transverse waves moving around the glass is:

v = 748 m/s (rounded to three significant figures)

Quiz 14: Sound

When the standing wave pattern in a pipe is NANA,the pipe has which of the following set of properties? (N stands for node,A for antinode. )

A standing wave is set up in a 200-cm string fixed at both ends.The string vibrates in 8 distinct segments when driven by a 120-Hz source.What is the wavelength?

The lower A on a piano has a frequency of 30.0 Hz.If the tension in the 2.00-m-long string is 316 N and one-half wavelength occupies the string,what is the mass of the string?

If a guitar string has a fundamental frequency of 512 Hz,what is the frequency of its first overtone?

A C note (f = 256 Hz) is sounded on a piano.If the length of the piano wire is 1.8 m and its mass density is 2.2 g/m,what is the tension in the wire?

What is the lowest frequency that will resonate in an organ pipe 5.00 m in length,closed at one end? The speed of sound in air is 340 m/s.

A tuning fork is sounded above a resonating tube (one end closed) ,which resonates at a length of 0.40 m and again at 1.2 m.If the tube length were extended further,at what point will the tuning fork again create a resonance condition?

A 2.5-m-long organ pipe is open at one end and closed at the other.Its fundamental tone has wavelength

A 1.30-m string is held fixed at both ends.When driven by a 140-Hz source,the string vibrates in 4 distinct segments.What is the natural fundamental frequency of the string?

An astronaut sets up a pendulum on the Moon,where the acceleration due to gravity is 1/6 that of Earth.If she finds that she must push the pendulum with a frequency of 0.10 Hz to maintain a swinging motion,what is the length of the pendulum?

A child sits on a swing supported by ropes of length 2.0 m.With what frequency will she need to apply the driving force to maintain swinging?

Quintupling the tension in a guitar string will change its natural frequency by what factor?

A pipe is open at both ends.The distance from a node to its adjacent antinode is equal to _______.

A 90-m-long high-voltage cable is suspended between two towers.The mass of the 90-m cable is 100 kg.If the tension in the cable is 25 000 N,what is the lowest frequency at which this cable can oscillate?

A string with fixed ends is made to go into standing wave patterns.At a given tension,the lowest frequency for which such a pattern exists is 150 Hz.Which of the following frequencies would also cause a standing wave pattern in this case?

A note is being played on a string on a guitar.The string is then pushed to the side while still in contact with the same fret.What happens to the frequency of the note then played compared to the original note?

If I quadruple the mass per unit length of guitar string,its natural frequency changes by what factor?

For a standing wave on a string the wavelength must equal

Standing-wave vibrations are set up in a crystal wine glass with four nodes and four antinodes equally spaced around the 2.80-cm radius of the rim.If the resonant frequency of the glass is 8.50 kHz,what is the speed of the transverse waves moving around the glass?

Quiz 14: Sound

When the standing wave pattern in a pipe is NANA,the pipe has which of the following set of properties? (N stands for node,A for antinode. )

A standing wave is set up in a 200-cm string fixed at both ends.The string vibrates in 8 distinct segments when driven by a 120-Hz source.What is the wavelength?

The lower A on a piano has a frequency of 30.0 Hz.If the tension in the 2.00-m-long string is 316 N and one-half wavelength occupies the string,what is the mass of the string?

If a guitar string has a fundamental frequency of 512 Hz,what is the frequency of its first overtone?

A C note (f = 256 Hz) is sounded on a piano.If the length of the piano wire is 1.8 m and its mass density is 2.2 g/m,what is the tension in the wire?

What is the lowest frequency that will resonate in an organ pipe 5.00 m in length,closed at one end? The speed of sound in air is 340 m/s.

A tuning fork is sounded above a resonating tube (one end closed) ,which resonates at a length of 0.40 m and again at 1.2 m.If the tube length were extended further,at what point will the tuning fork again create a resonance condition?

A 2.5-m-long organ pipe is open at one end and closed at the other.Its fundamental tone has wavelength

A 1.30-m string is held fixed at both ends.When driven by a 140-Hz source,the string vibrates in 4 distinct segments.What is the natural fundamental frequency of the string?

An astronaut sets up a pendulum on the Moon,where the acceleration due to gravity is 1/6 that of Earth.If she finds that she must push the pendulum with a frequency of 0.10 Hz to maintain a swinging motion,what is the length of the pendulum?

A child sits on a swing supported by ropes of length 2.0 m.With what frequency will she need to apply the driving force to maintain swinging?

Quintupling the tension in a guitar string will change its natural frequency by what factor?

A pipe is open at both ends.The distance from a node to its adjacent antinode is equal to _______.​

A 90-m-long high-voltage cable is suspended between two towers.The mass of the 90-m cable is 100 kg.If the tension in the cable is 25 000 N,what is the lowest frequency at which this cable can oscillate?

A string with fixed ends is made to go into standing wave patterns.At a given tension,the lowest frequency for which such a pattern exists is 150 Hz.Which of the following frequencies would also cause a standing wave pattern in this case?

A note is being played on a string on a guitar.The string is then pushed to the side while still in contact with the same fret.What happens to the frequency of the note then played compared to the original note?

If I quadruple the mass per unit length of guitar string,its natural frequency changes by what factor?

For a standing wave on a string the wavelength must equal

Standing-wave vibrations are set up in a crystal wine glass with four nodes and four antinodes equally spaced around the 2.80-cm radius of the rim.If the resonant frequency of the glass is 8.50 kHz,what is the speed of the transverse waves moving around the glass?

A pipe is open at both ends.The distance from a node to its adjacent antinode is equal to _______.