Apply Wave Model - IB Physics

A standing wave is established on a string of length $L = 0.60 \text{ m}$ fixed at both ends. The wave speed is $v = 240 \text{ m/s}$ . When the string vibrates in its third harmonic, what is the minimum time required for a particle initially at rest at an antinode to return to rest at the same position?
Answer: $2.50 \times 10^{-3} \text{ s}$ For the third harmonic, $f_3 = \frac{3v}{2L} = \frac{3 \times 240}{2 \times 0.60} = 600 \text{ Hz}$ . The period is $T = \frac{1}{f_3} = \frac{1}{600} = 5.0 \times 10^{-3} \text{ s}$ . A particle at an antinode oscillates with displacement $y(t) = A\cos(2\pi f_3 t)$ . Starting from rest at $t = 0$ means the particle is at maximum displacement. It returns to rest at the same position after half a period: $\frac{T}{2} = \frac{5.0 \times 10^{-3}}{2} = 2.50 \times 10^{-3} \text{ s}$ .
A sound wave travels through air as a series of compressions and rarefactions. What are the displacement and pressure of the air particles at the center of a compression?
Answer: The particle displacement is zero, and the pressure is at a maximum. A compression is a region where the pressure is at its maximum. Air particles move from regions of high pressure to low pressure. At the exact center of a compression, particles have moved in from both sides, and the net displacement of the particle at that equilibrium position is zero. The pressure, however, is at its peak.
An ultraviolet (UV) wave and an infrared (IR) wave are both travelling in a vacuum. Which statement correctly compares the UV wave to the IR wave?
Answer: The UV wave has the same speed and a shorter wavelength. Both UV and IR are electromagnetic waves, so they travel at the same speed (the speed of light, c) in a vacuum. In the electromagnetic spectrum, UV radiation has a higher frequency and thus a shorter wavelength than IR radiation, according to the relation $c = f\lambda$ . Therefore, the UV wave has the same speed and a shorter wavelength.
The intensity of a sound wave is proportional to the square of its displacement amplitude. A sound wave X has an intensity $I$ . A second sound wave Y, travelling in the same medium, has a displacement amplitude that is one-third the amplitude of wave X. What is the intensity of wave Y?
Answer: $I/9$ Let the amplitude of wave X be $A_X$ and its intensity be $I_X = I$ . We are given that $I \propto A^2$ . The amplitude of wave Y is $A_Y = A_X/3$ . The intensity of wave Y, $I_Y$ , will be proportional to $(A_Y)^2$ . Thus, the ratio of intensities is $I_Y / I_X = (A_Y)^2 / (A_X)^2 = (A_X/3)^2 / (A_X)^2 = (A_X^2/9) / A_X^2 = 1/9$ . Therefore, $I_Y = I_X/9 = I/9$ .
A sound wave is generated in air and then enters a large body of water. The speed of sound is greater in water than in air. Which property of the sound wave must remain constant as it crosses the boundary from air to water?
Answer: Frequency The frequency of a wave is determined by its source and does not change when the wave passes from one medium to another. The wave speed changes with the medium, as stated in the question. Since $v = f\lambda$ , a change in speed with constant frequency implies that the wavelength must also change. The amplitude will change as some of the wave's energy is reflected at the boundary and some is transmitted.
A source produces a wave of period $T$ that travels through a medium with speed $v$ . What is the shortest distance between two points on the wave that are oscillating with a phase difference of $\pi/2$ radians?
Answer: $vT/4$ One full wavelength, $\lambda$ , corresponds to a phase difference of $2\pi$ radians. The wavelength can be expressed as $\lambda = vT$ . A phase difference of $\pi/2$ is $(\pi/2) / (2\pi) = 1/4$ of a full cycle. Therefore, the distance between these two points is $1/4$ of a wavelength. The distance is $\lambda/4 = (vT)/4$ .
A radio station broadcasts at a frequency of 102.5 MHz. The waves travel from the transmitter to a receiver 60.0 km away. Approximately how many full wavelengths fit between the transmitter and the receiver? (Speed of light $c = 3.00 \times 10^8$ m s⁻¹)
Answer: 2.05 \times $10^4$ First, calculate the wavelength using $\lambda = v/f$ . The frequency is $f = 102.5 \times 10^6$ Hz and the speed is $v = c = 3.00 \times 10^8$ m s⁻¹. So, $\lambda = (3.00 \times 10^8) / (102.5 \times 10^6) \approx 2.927$ m. Next, calculate the number of wavelengths (N) in the distance ( $d = 60.0 km = 60.0 \times 10^3 m$ ). $N = d / \lambda = (60.0 \times 10^3) / 2.927 \approx 20499$ . This is approximately $2.05 \times 10^4$ .
Two sound waves, X and Y, travel through the same body of air at the same temperature. The wavelength of wave X is twice the wavelength of wave Y. What is the ratio of the speed of wave X to the speed of wave Y ( $v_X / v_Y$ )?
Answer: 1 The speed of a mechanical wave, such as sound, depends only on the properties of the medium through which it travels (in this case, the air's temperature, pressure, and density). Since both waves travel through the same medium, their speeds must be identical. Therefore, the ratio $v_X / v_Y$ is 1.
A transverse wave travels from left to right along a horizontal string. At a particular instant, a point P on the string is at the crest of the wave (maximum positive displacement). What is the instantaneous velocity of the particle at point P?
Answer: Zero The particles of the string oscillate in simple harmonic motion. At the point of maximum displacement (the crest), the particle momentarily stops before changing its direction to move downwards. Therefore, its instantaneous velocity is zero. The velocity of the wave is to the right, but this is different from the velocity of the individual particles of the medium.
An observer sees a flash of lightning and hears the corresponding thunder 4.5 s later. Assuming the speed of sound in air is 340 m s⁻¹, what is the best estimate for the distance to the lightning strike?
Answer: 1.5 km The speed of light is extremely high ( $3.0 \times 10^8$ m s⁻¹), so the time taken for the light to reach the observer is negligible. The distance can be calculated using the time it takes for the sound to travel. Using the formula distance = speed × time, we get $d = 340 \text{ m s⁻¹} \times 4.5 \text{ s} = 1530 \text{ m}$ , which is approximately 1.5 km.
A powerful explosion occurs on the surface of the Sun. Which of the following best explains why an observer on Earth can see the flash of light but cannot hear the sound of the explosion?
Answer: Light is an electromagnetic wave that can travel through a vacuum, while sound is a mechanical wave that requires a medium. The primary reason is the fundamental difference between electromagnetic waves (like light) and mechanical waves (like sound). The space between the Sun and Earth is a near-perfect vacuum. Electromagnetic waves do not require a medium for propagation and can travel through this vacuum. Mechanical waves require a medium (like air, water, or solids) to transfer energy through particle oscillations. Without a medium, sound cannot travel from the Sun to Earth.
Which phenomenon is characteristic of transverse waves but not of longitudinal waves?
Answer: Polarization Polarization is the restriction of the orientation of oscillations to a single plane. This is only possible for transverse waves, where the oscillations are perpendicular to the direction of energy transfer. Longitudinal waves, where oscillations are parallel to the direction of energy transfer, cannot be polarized. Refraction, diffraction, and superposition are phenomena exhibited by all types of waves.
A point on a transverse wave oscillates with a period $T$ . In a time interval equal to one-quarter of the period, the wave itself travels a distance $d$ . What is the wavelength of the wave?
Answer: $4d$ The speed of the wave is $v = \text{distance} / \text{time}$ . The time interval is $t = T/4$ and the distance is $d$ , so $v = d / (T/4) = 4d/T$ . The wavelength $\lambda$ is the distance the wave travels in one full period, so $\lambda = v \times T$ . Substituting the expression for $v$ , we get $\lambda = (4d/T) \times T = 4d$ .
A wave of frequency $f$ travels along a stretched string where its speed is $v$ . The tension in the string is then adjusted so that the wave speed becomes $2v$ . If the source continues to generate waves with the same frequency $f$ , what is the new wavelength?
Answer: $2\lambda$ The relationship between wave speed (v), frequency (f), and wavelength ( $\lambda$ ) is $v = f\lambda$ . Initially, $\lambda = v/f$ . The frequency $f$ is determined by the source and remains constant. The new speed is $v' = 2v$ . The new wavelength is $\lambda' = v'/f = (2v)/f = 2(v/f) = 2\lambda$ .
A wave pulse traveling on a rope encounters a junction where three identical ropes meet. The incident pulse has amplitude $A$ and carries power $P$ . Assuming no energy is lost at the junction and the transmitted pulses maintain their shape, what is the amplitude of each transmitted pulse?
Answer: $\frac{A}{\sqrt{3}}$ When a wave pulse splits at a junction, you need to apply two fundamental principles: conservation of energy and the relationship between wave power and amplitude. Since the three ropes are identical, symmetry tells us each transmitted pulse will have the same amplitude. The power carried by a wave is proportional to the square of its amplitude: $P \propto A^2$ . Since no energy is lost at the junction, the total power must be conserved. The incident pulse carries power $P$ , and this power splits equally among the three transmitted pulses, so each carries power $\frac{P}{3}$ . If each transmitted pulse has amplitude $A_t$ , then $\frac{P}{3} \propto A_t^2$ . Since the original power satisfies $P \propto A^2$ , we can write: $\frac{P}{3} = \frac{A^2}{3} \times \text{(proportionality constant)}$ Therefore: $A_t^2 = \frac{A^2}{3}$ , which gives $A_t = \frac{A}{\sqrt{3}}$ . Answer A ( $\frac{A\sqrt{3}}{3}$ ) is actually equivalent to the correct answer when rationalized, but it's written in an unnecessarily complex form. Answer B ( $\frac{A}{2}$ ) incorrectly assumes the amplitude scales linearly with the number of branches. Answer C ( $\frac{A}{3}$ ) makes the same linear scaling error, dividing amplitude directly by three instead of considering that power depends on amplitude squared. Remember: when waves split symmetrically, power divides equally among branches, but since power depends on amplitude squared, you must take the square root when finding the new amplitude.
A wave undergoes total internal reflection at the boundary between two media. The incident wave has amplitude $A_i$ and the reflected wave has amplitude $A_r$ . Due to the phase change upon reflection, a standing wave pattern forms in the incident medium. What is the amplitude of oscillation at a point located $\lambda/8$ from the boundary?
Answer: $A_i\sqrt{2}$ For total internal reflection, $|A_r| = |A_i|$ but with a phase change of $\pi$ . The standing wave amplitude at distance $x$ from the boundary is $A(x) = 2A_i|\sin(kx)|$ where $k = 2\pi/\lambda$ . At $x = \lambda/8$ : $kx = 2\pi \cdot \frac{\lambda/8}{\lambda} = \frac{\pi}{4}$ . Therefore $A(\lambda/8) = 2A_i\sin(\pi/4) = 2A_i \cdot \frac{1}{\sqrt{2}} = A_i\sqrt{2}$ .
A monochromatic light wave travels from air (refractive index ≈ 1.0) into a block of glass (refractive index ≈ 1.5). Which row correctly describes the changes, if any, to the frequency, wavelength, and speed of the light wave?
Answer: The frequency is constant, the wavelength decreases, and the speed decreases. The frequency of a wave is determined by its source and does not change when the wave enters a new medium. The speed of light decreases in a denser medium (higher refractive index). According to the wave equation $v = f\lambda$ , if $v$ decreases and $f$ remains constant, the wavelength $\lambda$ must also decrease.
A wave is generated in a Slinky spring. An observer notes that the coils of the spring move back and forth along the same axis as the wave propagates. The distance between consecutive regions of maximum coil density is 0.80 m. The wave travels 4.0 m in 2.0 s. What are the type and frequency of the wave?
Answer: Longitudinal, 2.5 Hz The coils move parallel to the wave propagation, which defines a longitudinal wave. The distance between consecutive regions of maximum density (compressions) is the wavelength, so $\lambda = 0.80$ m. The speed of the wave is $v = \text{distance} / \text{time} = 4.0 \text{ m} / 2.0 \text{ s} = 2.0$ m s⁻¹. The frequency is found using $f = v / \lambda = 2.0 / 0.80 = 2.5$ Hz.
Two coherent wave sources $S_1$ and $S_2$ separated by distance $d = 4.0 \text{ m}$ emit waves in phase with wavelength $\lambda = 1.2 \text{ m}$ . A detector moves along a line parallel to the line connecting the sources, at distance $L = 12 \text{ m}$ from the midpoint. At what distance from the perpendicular bisector will the detector first encounter a minimum in the interference pattern?
Answer: $1.8 \text{ m}$ For destructive interference, the path difference must be $(m + \frac{1}{2})\lambda$ where $m$ is an integer. The path difference is $\Delta = \sqrt{L^2 + (x + d/2)^2} - \sqrt{L^2 + (x - d/2)^2}$ . For small angles, $\Delta \approx \frac{xd}{L}$ . For the first minimum ( $m = 0$ ): $\frac{xd}{L} = \frac{\lambda}{2}$ . Substituting: $x = \frac{\lambda L}{2d} = \frac{1.2 \times 12}{2 \times 4.0} = 1.8 \text{ m}$ .
A continuous transverse wave is travelling along a string. Two points on the string, P and Q, are separated by a distance equal to three-quarters of the wavelength ( $3\lambda/4$ ). What is the phase difference between the oscillations of points P and Q?
Answer: $3\pi/2$ radians A full wavelength ( $\lambda$ ) corresponds to a phase difference of $2\pi$ radians. The phase difference is directly proportional to the separation distance. Therefore, for a separation of $3\lambda/4$ , the phase difference $\Delta\phi$ is $(3\lambda/4) / \lambda \times 2\pi = (3/4) \times 2\pi = 3\pi/2$ radians.

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