A mechanical system is oscillating at its natural frequency. If the system is subjected to light damping, which statement correctly describes the primary effects on the system's oscillations?
Answer: The amplitude decreases exponentially, while the frequency of oscillation remains approximately constant.
Light damping causes the amplitude of oscillations to decrease exponentially with time due to energy dissipation. However, the frequency of the oscillations is only slightly reduced from the natural frequency and can be considered approximately constant for light damping. Heavy or critical damping would cause oscillations to cease much more quickly.
Two points on a string support a standing wave. One point is at an antinode, and the other is in an adjacent segment, also at an antinode. What is the phase difference between the oscillations of these two points?
Answer: π radians
In a standing wave, all points within a single segment (between two consecutive nodes) oscillate in phase (phase difference of 0). However, points in adjacent segments are always out of phase. This means that when one segment is moving up, the next one is moving down. This corresponds to a phase difference of π radians (or 180°).
To achieve the largest possible amplitude when pushing a child on a swing, the pushes should be applied at a frequency that matches the natural frequency of the swing. This phenomenon is known as:
Answer: Resonance
Resonance is the phenomenon where a driving frequency matches the natural frequency of an oscillating system, leading to a dramatic increase in amplitude. Pushing a swing at its natural frequency is a classic example of resonance, as it maximizes the transfer of energy to the swing.
A pipe closed at one end can only produce standing waves at odd harmonics (1st, 3rd, 5th, etc.). What is the fundamental physical reason for this constraint?
Answer: The boundary conditions require a displacement node at the closed end and a displacement antinode at the open end.
The formation of a standing wave is governed by the boundary conditions. For a pipe closed at one end, the air molecules cannot oscillate longitudinally at the closed end, creating a displacement node. At the open end, the molecules have maximum freedom to oscillate, creating a displacement antinode. This combination of a node at one end and an antinode at the other can only be satisfied if the pipe's length is an odd multiple of quarter-wavelengths (L = nλ/4 for n=1, 3, 5...), which gives rise to only odd harmonics.
A car's suspension system is designed to be critically damped. If the system were instead heavily overdamped, what would be the most likely behaviour of the car after hitting a large bump?
Answer: The car would return to its equilibrium level very slowly, without any oscillation.
Critical damping provides the fastest return to equilibrium without oscillation. Overdamping (or heavy damping) also prevents oscillation, but the restoring force is so sluggish that the system takes a much longer time to return to equilibrium. Underdamping would cause the car to oscillate.
A string of length 0.80 m is fixed at both ends and vibrates in its fundamental mode. The speed of the wave on the string is 200 m s⁻¹. The sound produced travels through air, where the speed of sound is 340 m s⁻¹. What is the wavelength of the sound wave in the air?
Answer: 2.7 m
First, find the frequency of the string. In the fundamental mode (n=1), the wavelength on the string is λ_s = 2L = 2 * 0.80 m = 1.6 m. The frequency of vibration is f = vs / λ_s = 200 m s⁻¹ / 1.6 m = 125 Hz. The sound wave in the air has the same frequency as the source (the string). Second, use this frequency to find the wavelength in air: λ_air = v_air / f = 340 m s⁻¹ / 125 Hz = 2.72 m, which is approximately 2.7 m.
A string of length Ls vibrates in its second harmonic. The sound produced causes resonance in a pipe of length Lp, which is open at both ends and vibrating in its fundamental mode. The speed of waves on the string is vs and the speed of sound in air is va. What is the ratio Ls / Lp?
Answer: 2vs / va
For the string in its 2nd harmonic, Ls = 2(λs/2) = λs. The frequency is f = vs/λs = vs/Ls. For the open pipe in its fundamental mode, Lp = λa/2, so λa = 2Lp. The frequency is f = va/λa = va/(2Lp). For resonance, the frequencies must be equal: vs/Ls = va/(2Lp). Rearranging for the ratio Ls/Lp gives Ls/Lp = 2vs/va.
A musician playing a brass instrument, which can be modelled as a pipe, blows warmer air into it during a performance. Assuming the length of the instrument remains constant, what is the effect on the fundamental frequency of the notes produced?
Answer: The frequency increases because the speed of sound increases.
The speed of sound in a gas is proportional to the square root of its absolute temperature (v ∝ √T). Warmer air means a higher temperature, which increases the speed of sound v. The fundamental frequency of a pipe is proportional to the speed of sound (e.g., f = v/2L or f = v/4L). Since L is constant and v increases, the frequency f must increase, resulting in a higher pitch.
A resonant system with a very low degree of damping is described as having a high Quality factor (Q-factor). What is a characteristic of such a system?
Answer: It responds with a large amplitude over a very narrow range of driving frequencies.
A high Q-factor implies low damping. In a resonant system, low damping means that energy is dissipated slowly. When driven, this allows the amplitude to build up to a very large value, but only when the driving frequency is extremely close to the natural frequency. This results in a tall, sharp, and narrow resonance peak.
An organ pipe is open at both ends. Its fundamental frequency is f. What are the frequencies of the next two possible harmonics?
Answer: 2f, 3f
For a pipe open at both ends (or a string fixed at both ends), all integer harmonics are possible. The fundamental frequency corresponds to the first harmonic (n=1). The next two possible harmonics are the second (n=2) and third (n=3) harmonics. Their frequencies are integer multiples of the fundamental frequency, so they are 2f and 3f.
An experiment measures the amplitude of a driven oscillator as the driving frequency is varied, producing a resonance curve. If the experiment is repeated with significantly more damping, how will the new resonance curve compare to the original?
Answer: The peak amplitude will be lower and the curve will be broader.
Increased damping dissipates energy from the oscillating system more quickly. This has two main effects on the resonance curve: it reduces the maximum amplitude achieved at the resonant frequency, and it broadens the peak, meaning the system responds with significant amplitude over a wider range of driving frequencies.
Consider the energy distribution in a standing wave on a string. Which statement is correct?
Answer: Energy is stored within the segments between nodes but is not transferred along the string.
A key feature of a standing wave is that there is no net transfer of energy along the medium. Instead, energy is 'trapped' or stored in the segments between adjacent nodes. Within each segment, energy oscillates between kinetic energy (when the string passes through the equilibrium position) and potential energy (at maximum displacement).
Which statement correctly compares a standing wave and a progressive wave of the same frequency?
Answer: The phase difference between any two points on the standing wave is either 0 or π, whereas for a progressive wave it depends on their separation.
In a standing wave, all points within a segment oscillate in phase (phase difference 0), and points in adjacent segments oscillate in antiphase (phase difference π). In a progressive wave, the phase changes continuously with distance, so the phase difference between two points can be any value depending on their separation distance. This is a key distinction between the two wave types.
A vertical pipe of length 0.50 m is closed at its lower end. It resonates when a tuning fork with a frequency of 495 Hz is held at the open end. The speed of sound in air is 330 m s⁻¹. Which harmonic is being produced in the pipe?
Answer: Third harmonic
For a pipe closed at one end, the resonant frequencies are given by fn = n(v / 4L), where n must be an odd integer (1, 3, 5, ...). The fundamental frequency (n=1) is f1 = 1 (330 / (4 0.50)) = 165 Hz. The possible harmonic frequencies are 165 Hz, 495 Hz (3 165), 825 Hz (5 165), etc. Since the given frequency is 495 Hz, which is 3 times the fundamental, it is the third harmonic.
A string fixed at both ends has a fundamental frequency of 440 Hz. When the tension is increased by a factor of 1.44, what is the frequency of the third harmonic of the new configuration?
Answer: 1584 Hz
The fundamental frequency is proportional to the square root of tension: f∝T. When tension increases by 1.44, the new fundamental frequency is f′=440×1.44=440×1.2=528 Hz. The third harmonic is 3×528=1584 Hz. Choice B assumes linear relationship with tension. Choice C uses the original third harmonic. Choice D incorrectly calculates the fourth harmonic.
An organ pipe open at both ends resonates at 256 Hz for its fundamental mode. If one end is now closed, what will be the frequency of the first overtone of the modified pipe?
Answer: 384 Hz
For an open pipe, f0=2Lv=256 Hz, so Lv=512 Hz. When one end is closed, the fundamental becomes f0′=4Lv=128 Hz, and harmonics are only odd multiples. The first overtone is the third harmonic: 3×128=384 Hz. Choice B assumes even harmonics exist. Choice C uses the original fundamental. Choice D ignores the change in boundary conditions.
A pipe of length L, open at both ends, resonates at its fundamental frequency fo. One end of the pipe is then sealed. What is the new fundamental frequency of the modified pipe?
Answer: fo / 2
For a pipe open at both ends, the fundamental mode has a wavelength λ_o = 2L, so fo = v / (2L), where v is the speed of sound. For a pipe closed at one end, the fundamental mode has a wavelength λ_c = 4L. The new fundamental frequency is fc = v / (4L). By comparing the two expressions, we can see that fc = (1/2) * (v / (2L)) = fo / 2.
An organ pipe, open at both ends, is filled with air and produces a fundamental frequency f. If the air is replaced by helium at the same temperature, for which the speed of sound is significantly higher, what will be the new fundamental frequency f'?
Answer: f' > f
The fundamental frequency of a pipe open at both ends is given by f = v/(2L), where v is the speed of sound and L is the pipe's length. Since the length L remains constant, the frequency is directly proportional to the speed of sound. As the speed of sound in helium is significantly higher than in air, the new fundamental frequency f' will be significantly greater than f.
A stable standing wave is formed on a string of length L, fixed at both ends. Which of the following relationships between L and the wavelength λ of the wave is NOT possible?
Answer: L = 1.25λ
For a string fixed at both ends, a stable standing wave can only form if the length L is an integer multiple of half-wavelengths: L = n(λ/2), where n = 1, 2, 3, ... . Let's check the options: A) L = λ → n=2 (possible). B) L = 2λ → n=4 (possible). C) L = 1.5λ = 3λ/2 → n=3 (possible). D) L = 1.25λ = 5λ/4 → n(λ/2) = 5λ/4 → n = 2.5, which is not an integer. Therefore, this relationship is not possible.
A guitar string of length L, mass per unit length μ, and tension T vibrates in its fundamental mode at frequency f. A second string of the same length and material is subjected to a tension of 4T. What is the frequency of the third harmonic on this second string?
Answer: 6f
The speed of a wave on a string is given by v = √(T/μ). The new tension is T' = 4T, so the new speed is v' = √(4T/μ) = 2√(T/μ) = 2v. The fundamental frequency is f = v/(2L). The new fundamental frequency is f' = v'/(2L) = 2v/(2L) = 2f. The frequency of the nth harmonic is n times the fundamental frequency. Therefore, the frequency of the third harmonic on the second string is 3f' = 3(2f) = 6f.
