Apply Standing Waves & Resonance - IB Physics

In a resonance tube experiment, a tuning fork of unknown frequency is used. The first resonance occurs when the air column is 20.5 cm long, and the second resonance occurs when it is 62.0 cm long. If the speed of sound is 344 m/s, what is the frequency of the tuning fork?
Answer: 415 Hz For consecutive resonances in a closed tube, the difference in lengths equals half a wavelength: $\frac{\lambda}{2} = 62.0 - 20.5 = 41.5 cm = 0.415 m$ . Therefore, $\lambda = 0.83 m$ . The frequency is $f = \frac{v}{\lambda} = \frac{344}{0.83} = 414.5 Hz ≈ 415 Hz$ . Choice B incorrectly uses the first resonance length directly in calculations. Choice C uses an approximation that ignores end correction effects. Choice D uses an incorrect relationship between consecutive resonances.
A pipe is open at one end and closed at the other. For standing waves of air in the pipe, which statement correctly describes the displacement of air molecules and the pressure variation at the two ends?
Answer: The closed end has a displacement node and the open end has a pressure node. At the closed end of the pipe, air molecules cannot move longitudinally, so this must be a point of zero displacement (a displacement node). At the open end, the air is free to move, resulting in maximum displacement (a displacement antinode). Pressure and displacement are out of phase by $\pi/2$ . A displacement node corresponds to a pressure antinode (maximum pressure variation), and a displacement antinode corresponds to a pressure node (pressure is constant atmospheric pressure). Therefore, the closed end is a displacement node (and pressure antinode), and the open end is a displacement antinode (and pressure node). Option B correctly identifies two of these conditions.
A pipe open at both ends resonates at its fundamental frequency in air (speed of sound ≈ 340 m s⁻¹). The pipe is then filled with helium, in which the speed of sound is approximately three times greater. What is the new fundamental frequency of the pipe?
Answer: It is approximately 3 times the original frequency. The fundamental frequency of a pipe open at both ends is given by $f = v/(2L)$ , where $v$ is the speed of sound in the gas inside the pipe and $L$ is the length of the pipe. This relationship shows that the frequency is directly proportional to the speed of sound ( $f \propto v$ ). If the air is replaced with helium, where the speed of sound is three times greater, the new fundamental frequency will also be three times greater than the original frequency.
A mechanical system with low damping has a natural frequency of oscillation $f_0$ . The system is driven by an external force with a slowly increasing frequency $f$ . Which statement best describes the amplitude of the system's oscillation as $f$ increases from a value much less than $f_0$ to a value much greater than $f_0$ ?
Answer: The amplitude is small for $f \ll f_0$ , rises to a sharp maximum at $f \approx f_0$ , and becomes small again for $f \gg f_0$ . This question describes the phenomenon of resonance. When a system is driven by a periodic force, its response amplitude depends on the driving frequency $f$ . The amplitude is relatively small when $f$ is far from the natural frequency $f_0$ . As $f$ approaches $f_0$ , the amplitude increases dramatically, reaching a peak at or very near $f_0$ . For a system with low damping, this peak is sharp and high. As $f$ increases beyond $f_0$ , the amplitude decreases again. Option C provides the most accurate physical description of this resonance curve.
A guitar string of length 65 cm is plucked and vibrates in its fundamental mode with frequency 330 Hz. When the guitarist presses the string at the 5th fret (13 cm from the nut), reducing the vibrating length to 52 cm, and simultaneously increases the tension by 21%, what is the new fundamental frequency?
Answer: 450 Hz The fundamental frequency is $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ . When length changes from 65 cm to 52 cm and tension increases by 21% (factor of 1.21), the new frequency is $f' = f \times \frac{L_{old}}{L_{new}} \times \sqrt{1.21} = 330 \times \frac{65}{52} \times 1.1 = 330 \times 1.25 \times 1.1 = 453.75 Hz ≈ 450 Hz$ . Choice B incorrectly applies the tension factor. Choice C uses an incorrect length ratio. Choice D applies the factors incorrectly by multiplying instead of using the proper relationships.
A standing wave pattern is established in a string of length 1.5 m fixed at both ends. The standing wave has 3 complete half-wavelengths along the string's length. If a small piece of paper placed at 0.75 m from one end remains stationary, what can be concluded about the wave?
Answer: The paper is at a node and the wavelength is 1.0 m For 3 half-wavelengths in 1.5 m: λ/2 = 1.5/3 = 0.5 m, so λ = 1.0 m. Since the paper remains stationary, it must be at a node. For a string fixed at both ends, nodes occur at x = nλ/2 from one end. With λ = 1.0 m, nodes are at 0, 0.5, 1.0, and 1.5 m. The paper at 0.75 m is at the center, which is indeed a node position for this pattern.
A pipe closed at one end has a fundamental resonant frequency of 110 Hz. Which of the following frequencies will also produce resonance in this pipe?
Answer: 330 Hz A pipe closed at one end and open at the other only supports odd harmonics. The resonant frequencies are given by $f_n = n f_1$ , where n = 1, 3, 5, ... Given the fundamental frequency $f_1$ = 110 Hz, the next possible resonant frequencies are for n=3 ( $3 \times 110 = 330$ Hz) and n=5 ( $5 \times 110 = 550$ Hz). Even multiples of the fundamental, such as 220 Hz and 440 Hz, do not produce standing waves in this type of pipe. From the options, 330 Hz is a possible resonant frequency. 550 Hz is also a resonant frequency, but the question implicitly asks for the next one, which is 330 Hz. Both B and D are possible, but IB questions usually list only one correct option. Let me re-evaluate. A better phrasing might be Which of the following is a possible resonant frequency for this pipe?. Let's assume this intent. Then both 330 and 550 are correct. To make it a valid question, I must offer only one correct choice. I will change option D to something else. Let's make it 500 Hz. Re-writing answer choices. A: 220Hz, B: 330Hz, C: 440Hz, D: 500Hz. Then B is uniquely correct. Updated explanation: A pipe closed at one end and open at the other only supports odd harmonics. The resonant frequencies are given by $f_n = n f_1$ , where n = 1, 3, 5, ... Given the fundamental frequency $f_1$ = 110 Hz, the possible resonant frequencies are 110 Hz, 330 Hz, 550 Hz, etc. Of the choices given, only 330 Hz is a valid resonant frequency. 220 Hz and 440 Hz are even harmonics, which cannot exist in a closed pipe.
A string of length 0.60 m is fixed at both ends. It vibrates in its third harmonic with a frequency of 150 Hz. What is the speed of the waves on the string?
Answer: 60 m s⁻¹ For a string fixed at both ends, the length L is related to the wavelength λ by L = nλ/2, where n is the harmonic number. For the third harmonic, n=3. So, 0.60 m = 3λ/2, which gives λ = (2 × 0.60 m) / 3 = 0.40 m. The wave speed v is given by v = fλ. Therefore, v = 150 Hz × 0.40 m = 60 m s⁻¹.
An organ pipe of length L, open at both ends, is sounding its fundamental frequency, $f_{open}$ . A cap is then placed over one of the ends, closing it. What is the new fundamental frequency, $f_{closed}$ , of the modified pipe?
Answer: $f_{open} / 2$ For an open-open pipe of length L, the fundamental mode corresponds to half a wavelength fitting in the pipe, so $L = \lambda/2$ . The frequency is $f_{open} = v/\lambda = v/(2L)$ . When one end is closed, the pipe becomes a closed-open pipe. Its fundamental mode corresponds to a quarter of a wavelength fitting in the pipe, so $L = \lambda'/4$ . The new fundamental frequency is $f_{closed} = v/\lambda' = v/(4L)$ . By comparing the two expressions, we can see that $f_{closed} = \frac{v}{4L} = \frac{1}{2} \left( \frac{v}{2L} \right) = \frac{1}{2} f_{open}$ .
A string fixed at both ends has a fundamental frequency (first harmonic) of $f_0$ . What is the frequency of its third overtone?
Answer: $4f_0$ For a string fixed at both ends, all integer harmonics are present. The fundamental frequency (1st harmonic) is $f_0$ . Overtones are the frequencies higher than the fundamental. The first overtone is the 2nd harmonic ( $2f_0$ ). The second overtone is the 3rd harmonic ( $3f_0$ ). Therefore, the third overtone is the 4th harmonic, which has a frequency of $4f_0$ .
A string fixed at both ends has a fundamental frequency of 220 Hz when the tension is 400 N. If the tension is increased to 900 N while keeping all other parameters constant, what is the frequency of the third harmonic of the string?
Answer: 990 Hz The fundamental frequency is proportional to the square root of tension: $f \propto \sqrt{T}$ . When tension increases from 400 N to 900 N, the new fundamental frequency becomes $f' = 220 \times \sqrt{\frac{900}{400}} = 220 \times 1.5 = 330 Hz$ . The third harmonic is three times the fundamental frequency: $3 \times 330 = 990 Hz$ . Choice A incorrectly uses the square root relationship for the harmonic calculation. Choice C gives only the new fundamental frequency. Choice D incorrectly applies the tension ratio directly to the original frequency.
A pipe of length L is open at both ends and resonates at its fundamental frequency $f_1$ . What are the frequencies of the next two higher resonant modes?
Answer: 2 $f_1$ and 3 $f_1$ For a pipe open at both ends, all integer harmonics are present. The resonant frequencies are given by $f_n = n f_1$ , where n = 1, 2, 3, ... The fundamental frequency corresponds to n=1. The next two higher resonant modes are the second harmonic (n=2) and the third harmonic (n=3). Their frequencies are therefore 2 $f_1$ and 3 $f_1$ .
Consider three systems of length L, all resonating at their fundamental frequency: I. A string fixed at both ends. II. An air column in a pipe open at both ends. III. An air column in a pipe closed at one end. Which list correctly ranks the wavelengths ( $\lambda$ ) of the fundamental modes from longest to shortest?
Answer: III > I = II We need to find the wavelength for the fundamental mode (n=1) in each case. I. String fixed at both ends: $L = \lambda_I/2 \implies \lambda_I = 2L$ . II. Pipe open at both ends: $L = \lambda_{II}/2 \implies \lambda_{II} = 2L$ . III. Pipe closed at one end: $L = \lambda_{III}/4 \implies \lambda_{III} = 4L$ . Comparing the wavelengths, we see that $\lambda_{III} = 4L$ is the longest, and $\lambda_I = \lambda_{II} = 2L$ . Therefore, the correct ranking from longest to shortest is III > I = II.
A string fixed at both ends vibrates in its fundamental mode with frequency $f$ . The tension in the string is then increased by a factor of 4, while its length and mass per unit length remain unchanged. What is the new fundamental frequency?
Answer: $2f$ The fundamental frequency of a string is given by $f = v/(2L)$ , where $v$ is the wave speed and $L$ is the length. The wave speed on the string is given by $v = \sqrt{T/\mu}$ , where $T$ is the tension and $\mu$ is the mass per unit length. Combining these gives $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ . This shows that frequency is proportional to the square root of the tension ( $f \propto \sqrt{T}$ ). If the tension $T$ is increased by a factor of 4, the new frequency $f'$ will be proportional to $\sqrt{4T} = 2\sqrt{T}$ . Therefore, the new frequency is $2f$ .
A string vibrates in a standing wave pattern. Consider two points, P and Q, on the string. P is located between the first and second nodes, and Q is located between the second and third nodes. What is the phase difference between the oscillations of P and Q?
Answer: $\pi$ In a standing wave, all points within a single loop (i.e., between two adjacent nodes) oscillate in phase with each other (phase difference of 0). However, points in adjacent loops are in antiphase, meaning they oscillate with a phase difference of $\pi$ radians (180°). Since P and Q are in adjacent loops, the phase difference between their motions is $\pi$ .
Two waves traveling in opposite directions on a string have equations $y_1 = 0.05\sin(4\pi t - 2\pi x)$ and $y_2 = 0.05\sin(4\pi t + 2\pi x)$ where distances are in meters and time in seconds. At what positions along the string do nodes occur?
Answer: $x = 0.25 + 0.5n$ m, where n = 0, 1, 2, ... The superposition gives $y = y_1 + y_2 = 0.05[\sin(4\pi t - 2\pi x) + \sin(4\pi t + 2\pi x)] = 0.1\sin(4\pi t)\cos(2\pi x)$ . Nodes occur where the amplitude is zero, i.e., where $\cos(2\pi x) = 0$ . This happens when $2\pi x = \frac{\pi}{2} + n\pi$ , giving $x = \frac{1}{4} + \frac{n}{2} = 0.25 + 0.5n$ meters. Choice A gives antinode positions. Choice B gives positions spaced by half wavelength but starting at x = 0. Choice C incorrectly calculates the phase relationship.
A string of length L is fixed at both ends and is vibrating in its third harmonic (n=3). At which positions are the antinodes located, measured from one end (x=0)?
Answer: At L/6, L/2, and 5L/6 only. For the third harmonic (n=3) on a string of length L, the wavelength is given by $L = 3\lambda/2$ , so $\lambda = 2L/3$ . Nodes are located at positions of zero displacement, which for n=3 are at x=0, x=L/3, x=2L/3, and x=L. Antinodes are located at points of maximum displacement, which are halfway between adjacent nodes. The first antinode is at x = L/6. The second is at x = L/2. The third is at x = 5L/6.
An object with a natural frequency $f_0$ is subjected to a periodic driving force of variable frequency $f$ . If the amount of damping in the system is increased, how does the resonance curve (amplitude vs. driving frequency) change?
Answer: The resonance peak becomes broader and the maximum amplitude decreases. Damping is a process that removes energy from an oscillating system. Increasing damping has two main effects on the resonance curve: 1) It reduces the maximum amplitude achieved at the resonant frequency. 2) It broadens the resonance peak, meaning the system responds with a significant amplitude over a wider range of driving frequencies. A sharper peak and higher amplitude (option D) are characteristic of a system with less damping.
A tube closed at one end produces resonance for a sound source of 450 Hz. The next highest frequency at which resonance is produced is 750 Hz. What is the fundamental frequency of the tube?
Answer: 150 Hz A tube closed at one end only supports odd harmonics (n=1, 3, 5, ...). The resonant frequencies are $f_n = n f_1$ . The given frequencies, 450 Hz and 750 Hz, must be two consecutive odd harmonics. The difference in frequency between consecutive resonant modes is $f_{n+2} - f_n = (n+2)f_1 - nf_1 = 2f_1$ . Therefore, $750 \text{ Hz} - 450 \text{ Hz} = 300 \text{ Hz} = 2f_1$ . Solving for the fundamental frequency gives $f_1 = 300 / 2 = 150$ Hz. We can check this: the 3rd harmonic would be $3 \times 150 = 450$ Hz, and the 5th harmonic would be $5 \times 150 = 750$ Hz, which matches the problem statement.
Which statement best describes the net transfer of energy in a standing wave?
Answer: Energy oscillates between nodes but is not transferred along the medium as a whole. A key difference between a travelling wave and a standing wave is energy propagation. A travelling wave transports energy from one point to another. In an ideal standing wave, there is no net transfer of energy along the medium. Instead, energy is localized between the nodes. Within each segment (between two nodes), the energy oscillates between being entirely potential energy (at maximum displacement) and entirely kinetic energy (as particles pass through equilibrium).

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