Resononace, Vibrating Strings, and Air Columns in Pipes Practice Questions — JAMB UTME

Question 1Question →

A tube closed at one end has a length of $0.85\text{ m}$ . If the speed of sound in air is $340\text{ m/s}$ , what is the fundamental frequency of the sound wave produced in the tube?

100\text{ Hz}

200\text{ Hz}

400\text{ Hz}

289\text{ Hz}

Show answer & explanation

Answer: $100\text{ Hz}$

Answer

100\text{ Hz}

For a pipe closed at one end, the fundamental frequency is given by

f = \frac{v}{4L}

. Substituting

v = 340\text{ m/s}

and

L = 0.85\text{ m}

yields

f = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100\text{ Hz}

.

Step-by-Step Solution

1

Determine the relationship between pipe length and wavelength for the fundamental mode of a closed pipe.

For a pipe closed at one end,

\lambda = 4L = 4 \times 0.85\text{ m} = 3.4\text{ m}

.

A closed tube forms a node at the closed end and an antinode at the open end, corresponding to one quarter of a full wavelength.

2

Calculate fundamental frequency using the wave equation

v = f\lambda

.

f = \frac{v}{\lambda} = \frac{340\text{ m/s}}{3.4\text{ m}} = 100\text{ Hz}

.

Frequency equals wave velocity divided by fundamental wavelength.

Key Concept

Fundamental frequency of a pipe closed at one end

Question 2Question →

In a resonance tube experiment using a tuning fork of constant frequency, the first two consecutive resonant lengths of the air column above the water level are measured to be $23.5\text{ cm}$ and $73.5\text{ cm}$ respectively. What is the end correction of the tube in centimeters?

Show answer & explanation

Answer: 1.5

Answer

The end correction of the tube is

1.5\text{ cm}

.

In a resonance tube closed at one end by water, consecutive resonances occur when the air column length increases by half a wavelength. Subtracting the first resonant length from the second gives

\frac{\lambda}{2} = 73.5\text{ cm} - 23.5\text{ cm} = 50.0\text{ cm}

, which yields

\lambda = 100.0\text{ cm}

and

\frac{\lambda}{4} = 25.0\text{ cm}

. The first resonance condition accounts for end correction through

L_1 + e = \frac{\lambda}{4}

. Substituting

L_1 = 23.5\text{ cm}

gives

e = 25.0\text{ cm} - 23.5\text{ cm} = 1.5\text{ cm}

.

Step-by-Step Solution

1

Determine the wavelength using consecutive resonant positions

$\frac{\lambda}{2} = L_2 - L_1 = 73.5\text{ cm} - 23.5\text{ cm} = 50.0\text{ cm}$, so $\lambda = 100.0\text{ cm}$

For a column closed at one end, consecutive resonances occur at intervals of half a wavelength.

2

Calculate the quarter-wavelength value

$\frac{\lambda}{4} = \frac{100.0\text{ cm}}{4} = 25.0\text{ cm}$

The fundamental mode position of the displacement antinode corresponds to a distance of one quarter-wavelength from the closed end.

3

Calculate the end correction

$e = \frac{\lambda}{4} - L_1 = 25.0\text{ cm} - 23.5\text{ cm} = 1.5\text{ cm}$

The effective length for the first resonance includes the physical length plus the end correction.

Key Concept

End Correction in Resonance Air Columns

Question 3Question →

Match each acoustic or vibrating system operating under boundary conditions on the left with its corresponding fundamental or harmonic frequency relationship on the right (where $v$ is sound speed in air, $T$ is string tension, $\mu$ is linear mass density, $L$ is length, and $r$ is internal pipe radius).

Click a left item, then click its matching right item

Show answer & explanation

Answer

Pipe closed at one end matches

f = \frac{v}{4L}

; Pipe open at both ends matches

f = \frac{v}{2L}

; Stretched string in second harmonic matches

f = \frac{1}{L}\sqrt{\frac{T}{\mu}}

; Pipe closed at one end with end-correction matches

f = \frac{v}{4(L + 0.6r)}

.

Each system is correctly matched based on wave mechanics boundary conditions: closed pipes produce quarter-wave fundamental modes (

\lambda = 4L

), open pipes produce half-wave fundamental modes (

\lambda = 2L

), the second harmonic of a string doubles the fundamental frequency

f_1 = \frac{1}{2L}\sqrt{T/\mu}

to yield

f_2 = \frac{1}{L}\sqrt{T/\mu}

, and end-correction increases the effective length of a closed pipe to

L + 0.6r

.

Step-by-Step Solution

1

Analyze boundary conditions for an ideal closed pipe

Displacement node at closed end, antinode at open end. Length

L = \frac{\lambda}{4} \Rightarrow \lambda = 4L

. Frequency

f = \frac{v}{\lambda} = \frac{v}{4L}

.

Determines the fundamental mode frequency formula for a closed pipe without end correction.

2

Analyze boundary conditions for an ideal open pipe

Displacement antinodes at both open ends. Length

L = \frac{\lambda}{2} \Rightarrow \lambda = 2L

. Frequency

f = \frac{v}{2L}

.

Determines the fundamental mode frequency formula for an open pipe.

3

Calculate the second harmonic frequency of a stretched string

For wave speed

c = \sqrt{\frac{T}{\mu}}

, fundamental

f_1 = \frac{c}{2L}

. Second harmonic is

f_2 = 2f_1 = 2\left(\frac{1}{2L}\sqrt{\frac{T}{\mu}}\right) = \frac{1}{L}\sqrt{\frac{T}{\mu}}

.

Determines the frequency of the first overtone / second harmonic for a vibrating string fixed at both ends.

4

Apply end correction to a closed pipe

End correction

e = 0.6r

adds to physical length

L

at the open top end, giving

L_{eff} = L + 0.6r

. Fundamental frequency is

f = \frac{v}{4L_{eff}} = \frac{v}{4(L + 0.6r)}

.

Accounts for the antinode extending slightly beyond the open end of a real tube.

Key Concept

Boundary conditions, standing waves, harmonics in strings and air columns, and end-correction in resonance pipes

Question 4Question →

A stretched string of length $0.5\text{ m}$ fixed at both ends vibrates in its fundamental mode. If the speed of transverse waves along the string is $200\text{ m/s}$ , calculate the fundamental frequency of the string in hertz.

Show answer & explanation

Answer: 200

Answer

The fundamental frequency of the vibrating string is

200\text{ Hz}

.

For a string fixed at both ends, the fundamental mode corresponds to a standing wave with half a wavelength spanning the length of the string (

L = \frac{\lambda}{2}

, or

\lambda = 2L

). Applying the wave relation

v = f\lambda

, the fundamental frequency is

f = \frac{v}{2L}

. Substituting

v = 200\text{ m/s}

and

L = 0.5\text{ m}

gives

f = \frac{200}{2(0.5)} = 200\text{ Hz}

.

Step-by-Step Solution

1

Identify the relationship between frequency, wave speed, and string length for the fundamental mode.

For a string fixed at both ends, the wavelength of the fundamental harmonic is

\lambda = 2L

, giving the frequency formula

f = \frac{v}{2L}

.

The fundamental standing wave pattern contains nodes at both fixed ends and a single antinode at the center.

2

Substitute the given numerical values into the formula.

f = \frac{200\text{ m/s}}{2 \times 0.5\text{ m}} = \frac{200}{1} = 200\text{ Hz}

.

Dividing the wave speed by twice the length of the string yields the frequency in hertz.

Key Concept

Fundamental frequency of a vibrating string fixed at both ends

Question 5Question →

An organ pipe open at both ends has a length of $0.60\text{ m}$ . The fundamental frequency of resonance for this pipe is observed to be equal to the frequency of the first overtone of a stretched wire of length $0.40\text{ m}$ fixed at both ends. If the linear mass density of the wire is $4.0 \times 10^{-4}\text{ kg/m}$ and the speed of sound in air is $330\text{ m/s}$ , what is the tension in the wire?

1.21\text{ N}

4.84\text{ N}

9.68\text{ N}

19.36\text{ N}

Show answer & explanation

Answer: $4.84\text{ N}$

Answer

The tension in the wire is

4.84\text{ N}

.

The fundamental frequency of the open pipe is

f = \frac{v}{2L} = \frac{330}{2(0.60)} = 275\text{ Hz}

. Since the wire vibrates in its first overtone (second harmonic,

n=2

), its frequency is

f = \frac{2 v_{\text{wire}}}{2 L_{\text{wire}}} = \frac{v_{\text{wire}}}{0.40} = 275\text{ Hz}

. This gives

v_{\text{wire}} = 110\text{ m/s}

. Using

v_{\text{wire}} = \sqrt{T/\mu}

, the tension is

T = \mu v_{\text{wire}}^2 = (4.0 \times 10^{-4})(110)^2 = 4.84\text{ N}

.

Step-by-Step Solution

1

Calculate the fundamental frequency of the pipe open at both ends.

f_{\text{pipe}} = \frac{v_{\text{air}}}{2 L_{\text{pipe}}} = \frac{330}{2 \times 0.60} = 275\text{ Hz}

.

An open pipe supports a fundamental wavelength of

\lambda = 2L_{\text{pipe}}

.

2

Express the frequency of the first overtone of the stretched wire and equate it to the pipe frequency.

f_{\text{wire, 2}} = \frac{2 v_{\text{wire}}}{2 L_{\text{wire}}} = \frac{v_{\text{wire}}}{L_{\text{wire}}} = 275\text{ Hz}

.

For a wire fixed at both ends, the first overtone corresponds to the second harmonic (

n = 2

).

3

Solve for the speed of the transverse wave on the wire.

v_{\text{wire}} = 275 \times 0.40 = 110\text{ m/s}

.

Rearranging

f = v_{\text{wire}} / L_{\text{wire}}

gives

v_{\text{wire}} = f \cdot L_{\text{wire}}

.

4

Calculate the tension in the wire using the wave speed formula.

T = \mu v_{\text{wire}}^2 = (4.0 \times 10^{-4}) \times 110^2 = 4.84\text{ N}

.

The speed of a transverse wave on a stretched string is given by

v = \sqrt{T / \mu}

.

Key Concept

Resonance between air columns and vibrating strings under distinct boundary conditions

Estimated Time:2m 0s

Question 6Question →

A uniform wire of length $0.60\text{ m}$ and linear mass density $4.0 \times 10^{-3}\text{ kg/m}$ is fixed at both ends under a tension of $360\text{ N}$ . When plucked, the wire vibrates in its second overtone. This frequency is found to be in resonance with the first overtone of an air column in a pipe closed at one end. Taking the speed of sound in air as $340\text{ m/s}$ , what is the length of the pipe?

0.17\text{ m}

0.34\text{ m}

0.51\text{ m}

0.68\text{ m}

Show answer & explanation

Answer: $0.34\text{ m}$

Answer

The length of the pipe is

0.34\text{ m}

.

The correct answer of

0.34\text{ m}

is obtained by finding the wave speed on the string (

300\text{ m/s}

), computing its 3rd harmonic frequency (

750\text{ Hz}

), and equating this to the 3rd harmonic frequency formula for a closed pipe (

f = \frac{3 v_a}{4 L_p}

).

Step-by-Step Solution

1

Calculate the speed of transverse waves on the stretched string.

v_s = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{360\text{ N}}{4.0 \times 10^{-3}\text{ kg/m}}} = \sqrt{90000} = 300\text{ m/s}

Wave speed on a stretched string depends on tension and linear mass density.

2

Determine the fundamental frequency and the 2nd overtone frequency of the string.

Fundamental frequency

f_{1,s} = \frac{v_s}{2 L_s} = \frac{300}{2(0.60)} = 250\text{ Hz}

. Second overtone is the 3rd harmonic (

n=3

), so

f = 3 \times 250 = 750\text{ Hz}

.

For a string fixed at both ends, harmonics are integer multiples of the fundamental, and the 2nd overtone corresponds to

n=3

.

3

Relate the resonance frequency to the length of the closed pipe.

First overtone of a closed pipe is its 3rd harmonic (

m=3

):

f = \frac{3 v_a}{4 L_p} \implies 750 = \frac{3(340)}{4 L_p}

.

Pipes closed at one end only produce odd harmonics (

1, 3, 5, \dots

), so the 1st overtone is

m=3

.

4

Solve for the pipe length

L_p

.

L_p = \frac{3(340)}{4(750)} = \frac{1020}{3000} = 0.34\text{ m}

Algebraic rearrangement yields the required physical length.

Key Concept

Harmonics and overtones in vibrating strings and closed air columns under resonance

Question 7Question →

Match each vibrating system mode on the left with the correct relationship between its standing wavelength ( $\lambda$ ) and length ( $L$ ) on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The mode descriptions match their standing wavelength expressions as follows: Pipe closed at one end in its first overtone matches

\lambda = \frac{4L}{3}

; Pipe open at both ends in its first overtone matches

\lambda = L

; Stretched string in its second overtone matches

\lambda = \frac{2L}{3}

; Pipe closed at one end in its fundamental mode matches

\lambda = 4L

.

Each pair correctly links the specified boundary condition and mode of vibration to its mathematical relationship between wavelength

\lambda

and physical length

L

.

Step-by-Step Solution

1

Identify boundary conditions and available harmonics for each vibrating system.

Closed pipes support odd harmonics only (

n = 1, 3, 5, \dots

) with

L = \frac{n\lambda}{4}

. Open pipes and fixed strings support all integer harmonics (

n = 1, 2, 3, \dots

) with

L = \frac{n\lambda}{2}

.

Boundary conditions constrain node and antinode positions, determining allowed harmonic modes.

2

Determine the specific harmonic number

n

corresponding to each specified overtone.

First overtone of closed pipe

\rightarrow n = 3

; First overtone of open pipe

\rightarrow n = 2

; Second overtone of fixed string

\rightarrow n = 3

; Fundamental of closed pipe

\rightarrow n = 1

.

Overtones are higher resonant modes above the fundamental frequency.

3

Solve for wavelength

\lambda

in terms of system length

L

for each item.

For

n = 3

(closed pipe):

L = \frac{3\lambda}{4} \Rightarrow \lambda = \frac{4L}{3}

. For

n = 2

(open pipe):

L = \lambda \Rightarrow \lambda = L

. For

n = 3

(fixed string):

L = \frac{3\lambda}{2} \Rightarrow \lambda = \frac{2L}{3}

. For

n = 1

(closed pipe):

L = \frac{\lambda}{4} \Rightarrow \lambda = 4L

.

Rearranging each expression establishes the correct matching pair.

Key Concept

Boundary conditions and harmonic wavelength relations in pipes and vibrating strings

Question 8Question →

A pipe of length $0.80\text{ m}$ is closed at one end and open at the other. If the speed of sound in air is $320\text{ m/s}$ , what is the frequency of its first overtone?

100\text{ Hz}

200\text{ Hz}

300\text{ Hz}

400\text{ Hz}

Show answer & explanation

Answer: $300\text{ Hz}$

Answer

The frequency of the first overtone is

300\text{ Hz}

.

For an air column closed at one end, standing wave resonance occurs only at odd harmonic frequencies given by

f_n = \frac{n v}{4L}

for

n = 1, 3, 5, \dots

. The fundamental frequency (

n = 1

) is

f_1 = \frac{320}{4 \times 0.80} = 100\text{ Hz}

. The first overtone is the very next resonant mode, which corresponds to the third harmonic (

n = 3

), giving

f_3 = 3 \times 100\text{ Hz} = 300\text{ Hz}

.

Step-by-Step Solution

1

Calculate the fundamental frequency of the closed pipe

f_1 = \frac{v}{4L} = \frac{320}{4 \times 0.80} = 100\text{ Hz}

For a pipe closed at one end, the fundamental wavelength is

\lambda_1 = 4L

.

2

Determine the harmonic number for the first overtone

First overtone = 3rd harmonic (

f_3 = 3 f_1

)

A pipe closed at one end produces only odd harmonics (

f_n = n f_1

where

n = 1, 3, 5, \dots

).

3

Compute the first overtone frequency

f_3 = 3 \times 100\text{ Hz} = 300\text{ Hz}

Multiplying the fundamental frequency by 3 yields the first overtone frequency.

Key Concept

Resonance and Harmonics in Closed Air Columns

Question 9Question →

A stretched string of length $0.50\text{ m}$ fixed at both ends vibrates in its third harmonic mode at a frequency of $450\text{ Hz}$ . What is the speed of the transverse wave along the string in $\text{m/s}$ ?

Show answer & explanation

Answer: 150

Answer

The speed of the transverse wave along the string is

150\text{ m/s}

.

For a string fixed at both ends, standing wave modes produce harmonics given by

f_n = \frac{n v}{2L}

. Given

n = 3

,

L = 0.50\text{ m}

, and

f_3 = 450\text{ Hz}

, substituting these into the equation yields

450 = \frac{3v}{2(0.50)} = 3v

, leading to

v = 150\text{ m/s}

.

Step-by-Step Solution

1

Identify the standing wave frequency equation for a string fixed at both ends.

The frequency of the

n

-th harmonic is

f_n = \frac{n v}{2L}

, where

n

is the harmonic number,

v

is the wave speed, and

L

is the string length.

Fixed ends require nodes at both boundaries, producing standing wave modes with wavelengths

\lambda_n = \frac{2L}{n}

.

2

Substitute the given physical quantities into the harmonic equation.

450 = \frac{3 \times v}{2 \times 0.50}

.

The question specifies

L = 0.50\text{ m}

, third harmonic mode (

n = 3

), and frequency

f_3 = 450\text{ Hz}

.

3

Solve for the wave speed

v

.

v = 150\text{ m/s}

.

Simplifying

2 \times 0.50 = 1.0

gives

3v = 450

, so

v = \frac{450}{3} = 150\text{ m/s}

.

Key Concept

Standing Waves and Harmonics in Vibrating Strings

Question 10Question →

A uniform string of length $0.50\text{ m}$ and mass $2.0\text{ g}$ is fixed at both ends under a tension of $90\text{ N}$ . When vibrating, the second harmonic of this string resonates with the first overtone of an air column in a pipe closed at one end. Assuming the speed of sound in air is $340\text{ m/s}$ , calculate the length of the pipe in meters.

Show answer & explanation

Answer: 0.85

Answer

The length of the pipe is

0.85\text{ m}

.

The wave speed on the string is computed from tension and linear mass density as $150\text{ m/s}$, yielding a second harmonic frequency of $300\text{ Hz}$. Equating this to the first overtone (third harmonic) frequency formula of a closed air pipe, $f = \frac{3v_{air}}{4L_p}$, yields an exact pipe length of $0.85\text{ m}$.

Step-by-Step Solution

1

Calculate the linear mass density ($\mu$) of the string

$\mu = \frac{m}{L_s} = \frac{0.0020\text{ kg}}{0.50\text{ m}} = 4.0 \times 10^{-3}\text{ kg/m}$

Mass must be converted to kilograms before determining mass per unit length.

2

Determine the wave speed ($v_s$) along the stretched string

$v_s = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{90\text{ N}}{4.0 \times 10^{-3}\text{ kg/m}}} = \sqrt{22500} = 150\text{ m/s}$

The velocity of a transverse wave on a string depends on tension and linear mass density.

3

Calculate the second harmonic frequency of the string ($f_{2,s}$)

$f_{2,s} = \frac{v_s}{L_s} = \frac{150\text{ m/s}}{0.50\text{ m}} = 300\text{ Hz}$

The fundamental frequency is $f_{1,s} = \frac{v_s}{2L_s} = 150\text{ Hz}$, so the second harmonic is twice the fundamental frequency.

4

Set up the resonance equation for the first overtone of a closed pipe

$f_{3,p} = \frac{3 v_{air}}{4 L_p} = 300\text{ Hz}$

A pipe closed at one end produces only odd harmonics, so the first overtone is the 3rd harmonic.

5

Solve for the length of the closed pipe ($L_p$)

$L_p = \frac{3 \times 340\text{ m/s}}{4 \times 300\text{ Hz}} = \frac{1020}{1200} = 0.85\text{ m}$

Rearranging the frequency formula gives the required air column length.

Key Concept

Coupled resonance between standing waves on strings and air columns in closed pipes

Question 11Question →

An open pipe of physical length $0.58\text{ m}$ and a pipe closed at one end emit sound at the same frequency when both are vibrating in their first overtone mode. If the end correction at each open end is $0.01\text{ m}$ , what is the physical length of the closed pipe?

0.29\text{ m}

0.43\text{ m}

0.44\text{ m}

0.45\text{ m}

Show answer & explanation

Answer: $0.44\text{ m}$

Answer

The physical length of the closed pipe is

0.44\text{ m}

.

The effective length of an open pipe with two open ends is

L_{1,\text{eff}} = 0.58 + 2(0.01) = 0.60\text{ m}

. The first overtone for an open pipe is its second harmonic (

n=2

), giving

f = \frac{v}{0.60}

. For a pipe closed at one end, the first overtone is its third harmonic (

m=3

), giving

f = \frac{3v}{4L_{2,\text{eff}}}

. Equating frequencies gives

L_{2,\text{eff}} = 0.45\text{ m}

. Subtracting the single end correction (

e = 0.01\text{ m}

) yields the physical length

0.44\text{ m}

.

Step-by-Step Solution

1

Calculate the effective length and first overtone frequency of the open pipe.

L_{1,\text{eff}} = L_1 + 2e = 0.58\text{ m} + 2(0.01\text{ m}) = 0.60\text{ m}

, so

f_{\text{open,1st overtone}} = \frac{2v}{2L_{1,\text{eff}}} = \frac{v}{0.60}

.

An open pipe has two open ends, so end correction is applied at both ends (

2e

). Its first overtone corresponds to the second harmonic (

n=2

).

2

Express the first overtone frequency of the closed pipe in terms of its effective length.

f_{\text{closed,1st overtone}} = \frac{3v}{4L_{2,\text{eff}}}

.

A pipe closed at one end produces only odd harmonics (

m=1, 3, 5, \dots

). The first overtone corresponds to the third harmonic (

m=3

).

3

Equate the two frequencies and solve for the effective length of the closed pipe.

\frac{v}{0.60} = \frac{3v}{4L_{2,\text{eff}}} \implies 4L_{2,\text{eff}} = 1.80\text{ m} \implies L_{2,\text{eff}} = 0.45\text{ m}

.

Both pipes emit sound at the same frequency in their first overtone modes.

4

Calculate the physical length of the closed pipe.

L_2 = L_{2,\text{eff}} - e = 0.45\text{ m} - 0.01\text{ m} = 0.44\text{ m}

.

A closed pipe has only one open end, so its effective length is

L_2 + e

.

Key Concept

Standing Waves and End Correction in Open and Closed Organ Pipes

Question 12Question →

Match each vibrating acoustic system setup on the left with the correct mathematical expression for its resonant frequency ( $f$ ) on the right, where $v$ is the speed of sound in air, $L$ is the physical length of the pipe or string, $e$ is the end correction per open end, $T$ is tension, and $\mu$ is linear mass density.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The fundamental mode of a pipe closed at one end with end correction matches

f = \frac{v}{4(L + e)}

; the fundamental mode of a stretched string matches

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

; the fundamental mode of a pipe open at both ends with end correction at both ends matches

f = \frac{v}{2(L + 2e)}

; and the first overtone of a closed pipe without end correction matches

f = \frac{3v}{4L}

.

Each setup corresponds directly to its derived wave equation: closed pipes produce fundamental frequency

f = \frac{v}{4(L+e)}

for one open end, open pipes produce

f = \frac{v}{2(L+2e)}

for two open ends, stretched strings depend on tension and mass per unit length as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

, and the first overtone of a closed pipe is its third harmonic

f = \frac{3v}{4L}

.

Step-by-Step Solution

1

Analyze boundary conditions and effective acoustic length for closed and open pipes.

A closed pipe has one displacement antinode at the open end and one node at the closed end, adding an effective end correction

e

to its physical length

L

(

L_{\text{eff}} = L + e

). An open pipe has two open ends, giving an effective length

L_{\text{eff}} = L + 2e

.

Air displacement antinodes occur slightly outside open pipe boundaries by a distance

e

per open end.

2

Derive the frequency formula for the fundamental mode of a closed pipe with end correction.

For the fundamental mode,

L + e = \frac{\lambda}{4} \implies \lambda = 4(L + e)

. Frequency

f = \frac{v}{\lambda} = \frac{v}{4(L + e)}

.

The distance between a node and an adjacent antinode is one-quarter of a wavelength.

3

Derive the fundamental frequency for a stretched string fixed at both ends.

L = \frac{\lambda}{2} \implies \lambda = 2L

. Using wave velocity

v = \sqrt{\frac{T}{\mu}}

,

f = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

.

Nodes exist at both fixed ends in a vibrating string, making the fundamental wavelength twice the length.

4

Derive the fundamental frequency of an open pipe considering both end corrections.

L + 2e = \frac{\lambda}{2} \implies \lambda = 2(L + 2e)

, so

f = \frac{v}{2(L + 2e)}

.

Antinodes occur at both open ends, placing half a wavelength within the effective acoustic length.

5

Determine the first overtone frequency for a closed pipe without end correction.

The first overtone is the third harmonic (

n = 3

), so

L = \frac{3\lambda}{4} \implies \lambda = \frac{4L}{3}

, which gives

f = \frac{3v}{4L}

.

Closed pipes support only odd integer multiples of the fundamental frequency.

Key Concept

Standing Waves and Resonance in Air Columns and Strings

Question 13Question →

Match each physical modification of a vibrating string or air pipe system on the left with its corresponding effect on the system's frequency on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Quadrupling tension corresponds to increasing frequency by a factor of 2; quadrupling linear mass density corresponds to reducing frequency to half; doubling tension corresponds to increasing frequency by a factor of

\sqrt{2}

; transitioning a closed pipe from fundamental mode to first overtone corresponds to increasing frequency by a factor of 3.

Each physical modification correctly maps to its quantitative outcome based on wave mechanics: string frequency scales with

\sqrt{T}

and

1/\sqrt{\mu}

, while closed pipe overtones follow odd harmonic multipliers (

1, 3, 5, \dots

).

Step-by-Step Solution

1

Examine the fundamental frequency formula for a stretched string under tension:

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

.

Frequency is directly proportional to

\sqrt{T}

and inversely proportional to

\sqrt{\mu}

.

This establishes how changes in tension and mass per unit length scale the fundamental frequency.

2

Calculate scaling factors for the string modifications.

Quadrupling

T

multiplies frequency by

\sqrt{4} = 2

. Quadrupling

\mu

multiplies frequency by

1/\sqrt{4} = 0.5

. Doubling

T

multiplies frequency by

\sqrt{2}

.

Applying square roots to the parameter change factors gives the resultant frequency change.

3

Analyze harmonic ratios for air columns in pipes closed at one end.

The fundamental mode frequency is

f_1 = \frac{v}{4L}

. The first overtone is the third harmonic (

f_3 = \frac{3v}{4L} = 3f_1

).

Closed air columns produce only odd harmonics (

n = 1, 3, 5, \dots

).

Key Concept

Parameter scaling of transverse waves on stretched strings and harmonic modes in closed air columns

Question 14Question →

A pipe closed at one end vibrates in its first overtone. An open pipe vibrating in its fundamental mode has a frequency equal to that of the closed pipe. Neglecting end corrections, what is the ratio of the length of the open pipe to the length of the closed pipe?

2 : 3

3 : 2

4 : 3

1 : 3

Show answer & explanation

Answer: $2 : 3$

Answer

The ratio of the length of the open pipe to the length of the closed pipe is

2 : 3

.

The first overtone of a closed pipe corresponds to its 3rd harmonic, giving a frequency of

f = \frac{3v}{4L_c}

. Equating this to the fundamental frequency of an open pipe (

f = \frac{v}{2L_o}

) yields

\frac{v}{2L_o} = \frac{3v}{4L_c}

, which simplifies to

\frac{L_o}{L_c} = \frac{2}{3}

or

2 : 3

.

Step-by-Step Solution

1

Write the frequency formula for the first overtone of the closed pipe.

For a closed pipe of length

L_c

, odd harmonics are produced (

n = 1, 3, 5, \dots

). The first overtone is the third harmonic (

n = 3

):

f_{\text{closed}} = \frac{3v}{4L_c}

Closed air columns produce only odd harmonics, where the fundamental is

n=1

and the first overtone is

n=3

.

2

Write the fundamental frequency formula for the open pipe.

For an open pipe of length

L_o

, all harmonics are produced (

m = 1, 2, 3, \dots

). The fundamental frequency (

m = 1

) is:

f_{\text{open}} = \frac{v}{2L_o}

Open air columns have antinodes at both ends, yielding a fundamental wavelength of

\lambda = 2L_o

.

3

Equate the two frequencies and solve for the ratio

\frac{L_o}{L_c}

.

\frac{v}{2L_o} = \frac{3v}{4L_c}

Cancel the speed of sound

v

from both sides:

\frac{1}{2L_o} = \frac{3}{4L_c}

Cross-multiply:

6L_o = 4L_c \implies \frac{L_o}{L_c} = \frac{4}{6} = \frac{2}{3}

The question states that the frequencies of the two pipe configurations are equal.

Key Concept

Harmonics in Open and Closed Air Columns

Question 15Question →

In a resonance tube experiment using a tuning fork of frequency $340\text{ Hz}$ , the first two consecutive resonance lengths of the air column closed at one end are $22\text{ cm}$ and $72\text{ cm}$ . If a pipe open at both ends with a physical length of $28\text{ cm}$ is operated in the same environment, what is its fundamental frequency when end corrections at both open ends are taken into account?

500\text{ Hz}

250\text{ Hz}

680\text{ Hz}

1000\text{ Hz}

Show answer & explanation

Answer: $500\text{ Hz}$

Answer

The fundamental frequency of the open pipe is

500\text{ Hz}

.

The difference between successive resonant lengths in the closed tube gives half a wavelength (

L_2 - L_1 = 0.50\text{ m} \implies \lambda = 1.00\text{ m}

). Using the tuning fork frequency

340\text{ Hz}

, the speed of sound is

340\text{ m/s}

. The end correction is

e = \lambda / 4 - L_1 = 0.25\text{ m} - 0.22\text{ m} = 0.03\text{ m}

. For a pipe open at both ends, end corrections apply at both openings, making the effective length

L_{\text{eff}} = 0.28\text{ m} + 2(0.03\text{ m}) = 0.34\text{ m}

. Its fundamental frequency is

f_0 = v / (2 L_{\text{eff}}) = 340 / (2 \times 0.34) = 500\text{ Hz}

.

Step-by-Step Solution

1

Determine the wavelength and speed of sound from the resonance tube data.

\lambda = 2(L_2 - L_1) = 2(0.72\text{ m} - 0.22\text{ m}) = 1.00\text{ m}

. Speed of sound

v = f \lambda = 340\text{ Hz} \times 1.00\text{ m} = 340\text{ m/s}

.

The distance between consecutive resonance positions in a closed pipe is equal to half a wavelength.

2

Calculate the end correction

e

of the tube.

L_1 + e = \frac{\lambda}{4} \implies 0.22\text{ m} + e = 0.25\text{ m} \implies e = 0.03\text{ m} = 3\text{ cm}

.

The first resonance of a pipe closed at one end occurs when the effective length equals one quarter of a wavelength.

3

Calculate the effective length

L_{\text{eff}}

of the open pipe.

L_{\text{eff}} = L + 2e = 28\text{ cm} + 2(3\text{ cm}) = 34\text{ cm} = 0.34\text{ m}

.

A pipe open at both ends requires an end-correction term added at each open boundary.

4

Calculate the fundamental frequency of the open pipe.

f_0 = \frac{v}{2 L_{\text{eff}}} = \frac{340\text{ m/s}}{2 \times 0.34\text{ m}} = 500\text{ Hz}

.

The fundamental wavelength of a pipe open at both ends is twice its effective length.

Key Concept

Resonance tube end correction and boundary conditions of open vs closed pipes

Estimated Time:2m 0s

Question 16Question →

A sonometer wire of length $0.80\text{ m}$ and mass $2.0\text{ g}$ is maintained under a tension of $100\text{ N}$ . What is the fundamental frequency of vibration of the wire?

125\text{ Hz}

250\text{ Hz}

62.5\text{ Hz}

12.5\text{ Hz}

Show answer & explanation

Answer: $125\text{ Hz}$

Answer

The fundamental frequency of vibration of the wire is

125\text{ Hz}

.

The linear mass density is

\mu = \frac{0.002\text{ kg}}{0.80\text{ m}} = 0.0025\text{ kg/m}

. The velocity of waves on the string is

v = \sqrt{\frac{100}{0.0025}} = 200\text{ m/s}

. The fundamental frequency is

f = \frac{v}{2L} = \frac{200}{2 \times 0.80} = 125\text{ Hz}

.

Step-by-Step Solution

1

Convert mass to kilograms and calculate linear mass density

(\mu)

.

m = 2.0\text{ g} = 0.002\text{ kg}

. Thus,

\mu = \frac{m}{L} = \frac{0.002\text{ kg}}{0.80\text{ m}} = 0.0025\text{ kg/m} = 2.5 \times 10^{-3}\text{ kg/m}

.

Standard SI units must be used for tension in Newtons and length in meters.

2

Calculate the speed of the transverse wave on the string

(v)

.

v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.0025}} = \sqrt{40000} = 200\text{ m/s}

.

Wave speed on a stretched string depends directly on tension and inversely on linear density.

3

Calculate the fundamental frequency

(f_1)

.

f_1 = \frac{v}{2L} = \frac{200}{2 \times 0.80} = \frac{200}{1.6} = 125\text{ Hz}

.

For a fixed string vibrating in its fundamental mode, the length equals half the wavelength (

\,L = \frac{\lambda}{2}\,

).

Key Concept

Fundamental frequency of a stretched string

Question 17Question →

A uniform wire of length $0.60\text{ m}$ fixed at both ends vibrates in its fundamental mode with a frequency of $150\text{ Hz}$ . If the length of the wire is reduced to $0.40\text{ m}$ while the tension in the wire is increased by a factor of 4, what is the new fundamental frequency of vibration?

200\text{ Hz}

300\text{ Hz}

450\text{ Hz}

900\text{ Hz}

Show answer & explanation

Answer: $450\text{ Hz}$

Answer

The new fundamental frequency of the vibrating wire is

450\text{ Hz}

.

The fundamental frequency of a stretched string is inversely proportional to its length and directly proportional to the square root of its tension (

f \propto \frac{\sqrt{T}}{L}

). Reducing the length from

0.60\text{ m}

to

0.40\text{ m}

increases frequency by a factor of

\frac{0.60}{0.40} = 1.5

. Quadrupling the tension increases frequency by a factor of

\sqrt{4} = 2

. Combining both effects gives an overall frequency multiplier of

1.5 \times 2 = 3.0

, leading to

3.0 \times 150\text{ Hz} = 450\text{ Hz}

.

Step-by-Step Solution

1

Write the general formula for the fundamental frequency of a stretched string.

f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}

, where

L

is length,

T

is tension, and

\mu

is linear density.

Establishes the functional dependence of frequency on length and tension.

2

Set up the ratio between the new frequency

f_2

and initial frequency

f_1

.

\frac{f_2}{f_1} = \frac{L_1}{L_2} \sqrt{\frac{T_2}{T_1}}

.

Since linear mass density

\mu

remains constant, comparing ratios isolates the changing variables.

3

Substitute the given parameters into the ratio equation.

\frac{f_2}{150} = \frac{0.60}{0.40} \times \sqrt{4} = 1.5 \times 2 = 3.0

.

Calculates the scaling factor for the new fundamental frequency.

4

Solve for the new fundamental frequency

f_2

.

f_2 = 3.0 \times 150 = 450\text{ Hz}

.

Yields the final numerical value of the modified fundamental frequency.

Key Concept

Fundamental frequency of vibrating stretched strings under varying length and tension

Question 18Question →

Match each vibrating acoustic system setup on the left with its corresponding displacement standing wave characteristics (number of nodes, antinodes, and wavelength $\lambda$ in terms of pipe or string length $L$ ) on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Pipe closed at fundamental matches 1 node, 1 antinode ($\lambda = 4L$); Pipe open at fundamental matches 1 node, 2 antinodes ($\lambda = 2L$); String fixed at second harmonic matches 3 nodes, 2 antinodes ($\lambda = L$); Pipe closed at first overtone matches 2 nodes, 2 antinodes ($\lambda = \frac{4}{3}L$).

Each standing wave profile is uniquely determined by boundary constraints: fixed ends and closed pipe ends form displacement nodes, whereas open pipe ends form displacement antinodes. Counting the number of quarter-wavelength segments yields the exact relationship between wavelength $\lambda$ and system length $L$.

Step-by-Step Solution

1

Identify boundary conditions for displacement standing waves

Fixed ends of strings and closed ends of pipes are displacement nodes. Open ends of pipes are displacement antinodes.

Physical constraints prevent particle displacement at rigid boundaries while allowing maximum oscillation amplitude at open boundaries.

2

Calculate node/antinode count and wavelength for fundamental modes

For a closed pipe fundamental, $L = \frac{\lambda}{4}$, so $\lambda = 4L$ (1 node, 1 antinode). For an open pipe fundamental, $L = \frac{\lambda}{2}$, so $\lambda = 2L$ (1 node, 2 antinodes).

The distance between a consecutive node and antinode is $\frac{\lambda}{4}$, while the distance between two consecutive antinodes is $\frac{\lambda}{2}$.

3

Calculate node/antinode count and wavelength for higher harmonics

For a string fixed at both ends in the 2nd harmonic, two complete half-wavelength loops exist ($L = \lambda$), giving 3 nodes and 2 antinodes. For a closed pipe in its first overtone (3rd harmonic), $L = \frac{3\lambda}{4}$, so $\lambda = \frac{4}{3}L$, giving 2 nodes and 2 antinodes.

The harmonic number dictates how many quarter-wavelength or half-wavelength segments fit within length $L$.

Key Concept

Boundary conditions, node-antinode distributions, and wavelength formulas for standing waves in strings and pipes

Question 19Question →

A pipe closed at one end vibrates in its fundamental mode with a frequency equal to that of an open pipe vibrating in its first overtone. What is the ratio of the length of the closed pipe to the length of the open pipe?

1 : 4

1 : 2

3 : 4

4 : 1

Show answer & explanation

Answer: $1 : 4$

Answer

The ratio of the length of the closed pipe to the length of the open pipe is

1 : 4

.

The fundamental mode of a pipe closed at one end has a frequency of

f = \frac{v}{4L_c}

. The first overtone of a pipe open at both ends corresponds to the second harmonic, which has a frequency of

f = \frac{v}{L_o}

. Equating the two frequencies gives

\frac{v}{4L_c} = \frac{v}{L_o}

, which simplifies directly to

\frac{L_c}{L_o} = \frac{1}{4}

or

1 : 4

.

Step-by-Step Solution

1

Express the fundamental frequency of the pipe closed at one end

f_c = \frac{v}{4L_c}

, where

v

is the speed of sound and

L_c

is the length of the closed pipe.

A pipe closed at one end supports odd harmonics, and its fundamental wavelength is

\lambda_c = 4L_c

.

2

Express the frequency of the first overtone of the open pipe

f_o = \frac{2v}{2L_o} = \frac{v}{L_o}

, where

L_o

is the length of the open pipe.

An open pipe supports all harmonics (

n = 1, 2, 3, \dots

). The fundamental is

n=1

and the first overtone corresponds to

n=2

.

3

Equate the two frequencies and solve for the length ratio

\frac{L_c}{L_o}

\frac{v}{4L_c} = \frac{v}{L_o} \implies 4L_c = L_o \implies \frac{L_c}{L_o} = \frac{1}{4}

The problem states that the fundamental frequency of the closed pipe is equal to the first overtone frequency of the open pipe.

Key Concept

Standing Waves and Harmonics in Pipes

Resononace, Vibrating Strings, and Air Columns in Pipes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes