Waves and Oscillations

This problem involves fiber optic pulse transmission and signal processing.

a) Reasons for pulse spreading:

Two main reasons why a light pulse spreads out in a fiber:

1. Modal dispersion: Different light modes (paths) travel at different speeds through the fiber. Light entering at different angles takes different paths down the fiber, arriving at different times.

2. Chromatic dispersion: The refractive index of the fiber core depends on wavelength. Different colors in the pulse travel at different speeds, causing the pulse to spread.

b) Travel time range: Slowest pulse (minimum speed): $$t_{slow} = \frac{L}{v_{min}} = \frac{10.0 \times 10^3}{1.95 \times 10^8} = 5.128 \times 10^{-5} \text{ s}$$ Fastest pulse (maximum speed): $$t_{fast} = \frac{L}{v_{max}} = \frac{10.0 \times 10^3}{2.05 \times 10^8} = 4.878 \times 10^{-5} \text{ s}$$ c) Maximum LED frequency:

The pulse spreads over time by: $\Delta t = 5.128 \times 10^{-5} - 4.878 \times 10^{-5} = 0.25 \times 10^{-5}$ s

For non-overlapping pulses, the period must be at least twice the spread: $$T_{min} = 2\Delta t = 0.50 \times 10^{-5} \text{ s}$$

$$f_{max} = \frac{1}{T_{min}} = \frac{1}{0.50 \times 10^{-5}} = 2 \times 10^5 \text{ Hz} = 200 \text{ kHz}$$

d) Frequency of 1310 nm light:

$$c = f\lambda \Rightarrow f = \frac{c}{\lambda}$$

$$f = \frac{3.0 \times 10^8}{1310 \times 10^{-9}} = \frac{3.0 \times 10^8}{1.31 \times 10^{-6}}$$

$$f = 2.29 \times 10^{14} \text{ Hz} \approx 2.3 \times 10^{14} \text{ Hz}$$

e) Spectrum identification:

Given: Red light frequency $f_{red} = 4 \times 10^{14}$ Hz

Our frequency: $f = 2.3 \times 10^{14}$ Hz

Since $2.3 \times 10^{14} < 4 \times 10^{14}$, our light has lower frequency than red light.

Lower frequency means longer wavelength than red light.

Therefore, 1310 nm light is in the infrared region of the spectrum (beyond visible red light).

This problem involves wave physics, specifically gravity waves on water and their behavior in different depth conditions.

Given: - Deep water (h $\gg$ $\lambda$): $v = \sqrt{\frac{g\lambda}{2\pi}}$
- Shallow water (h $\ll$ $\lambda$): $v = \sqrt{gh}$
- Tsunami wavelength: $\lambda = 80$ km
- Ocean depth: $h_1 = 4.7$ km
- Coastal depth: $h_2 = 10$ m
- Ocean amplitude: $A_1 = 35$ cm
- $g = 9.8 m/s^2$
a) Hull speed derivation:

For a ship creating waves with wavelength $\lambda$ equal to ship length $L$:

In deep water: $v = \sqrt{\frac{g\lambda}{2\pi}}$

Substituting $\lambda = L$: $$v_{hull} = \sqrt{\frac{gL}{2\pi}}$$

$$v_{hull} = \sqrt{\frac{9.8 \times L}{2\pi}} = \sqrt{\frac{9.8}{2\pi}} \sqrt{L}$$

$$v_{hull} = \sqrt{1.56} \sqrt{L} \approx 1.25\sqrt{L} \approx 1.2L^{1/2}$$

b) Why $L$ must be in metres:

The formula $v_{hull} = 1.2\sqrt{L}$ depends on: $$1.2 = \sqrt{\frac{g}{2\pi}} = \sqrt{\frac{9.8}{2\pi}}$$

The value $g = 9.8$ is in SI units ($m/s^2$). For consistency:
- If $g$ is in $m/s^2$, then $L$ must be in metres
- The resulting velocity is in $m/s$
If $L$ were in other units (feet, km), the numerical factor 1.2 would change accordingly.

c) Wave speed in terms of period $T$:

Wave speed: $v = \frac{\lambda}{T}$

Therefore: $\lambda = vT$

Substituting into deep water formula: $$v = \sqrt{\frac{g(vT)}{2\pi}}$$

$$v^2 = \frac{gvT}{2\pi}$$

$$v = \frac{gT}{2\pi}$$

d) Tsunami speed calculations:

Tsunami wavelength $\lambda = 80$ km = 80,000 m

In deep ocean (h = 4.7 km = 4700 m):

Since $h = 4700 \ll \lambda = 80000$, this is shallow water: $$v = \sqrt{gh} = \sqrt{9.8 \times 4700}$$

$$v = \sqrt{46060} \approx 215 \text{ m/s}$$

In coastal shallows (h = 10 m):

$$v = \sqrt{9.8 \times 10} = \sqrt{98} \approx 9.9 \text{ m/s}$$

The tsunami slows dramatically from 215 m/s to 9.9 m/s as it approaches shore.

e) Amplitude-depth relationship:

Power of wave: $P \propto vA^2$

For energy conservation (constant power flow): $$vA^2 = \text{constant}$$

$$A^2 \propto \frac{1}{v}$$

In shallow water: $v = \sqrt{gh}$

$$A^2 \propto \frac{1}{\sqrt{h}} = h^{-1/2}$$

$$A \propto h^{-1/4}$$

This means as depth decreases, amplitude increases (but only as the fourth root).

f) Amplitude at 10 m depth:

$$A \propto h^{-1/4} \Rightarrow \frac{A_2}{A_1} = \left(\frac{h_2}{h_1}\right)^{-1/4}$$

$$A_2 = A_1 \left(\frac{h_2}{h_1}\right)^{-1/4} = 35 \times \left(\frac{10}{4700}\right)^{-1/4}$$

$$A_2 = 35 \times (0.00213)^{-1/4}$$

$$A_2 = 35 \times \frac{1}{(0.00213)^{1/4}} = 35 \times \frac{1}{0.217}$$

$$A_2 = 35 \times 4.61 \approx 160 \text{ cm}$$

The tsunami amplitude increases from 35 cm in deep ocean to about 160 cm (1.6 m) in shallow water, demonstrating why tsunamis become dangerous near coastlines.

This problem involves the reflection of light from a spherical mirror.

Understanding the geometry:

We have:
- Point light source embedded in a screen
- Circular mirror of diameter 30 cm (radius $r = 15$ cm)
- Mirror placed 20 cm in front of the screen
- Mirror center lies on the normal line from the light source
a) Path of light rays:

Light rays from the point source: 1. Travel outward in all directions 2. Some rays hit the mirror 3. Law of reflection: Angle of incidence = Angle of reflection 4. Reflected rays return to the screen

Key insight: For a point source on the axis of a spherical mirror, all reflected rays appear to diverge from a single point. b) Calculating the area of illumination: Understanding the image formation:

When light from a point source reflects off a spherical mirror:
- The reflected rays diverge (they don't converge to a point)
- They create a circle of light on the screen
- The diameter of this circle is twice the diameter of the mirror
Why twice the diameter?

Consider a ray from the source to the edge of the mirror:
- It hits the mirror at distance $r = 15$ cm from the axis
- Upon reflection, this ray travels back to the screen
- Due to the geometry, it lands at distance $2r = 30$ cm from the axis
Area calculation:

Radius of illuminated circle: $R = 30$ cm $= 0.30$ m

$$\text{Area} = \pi R^2 = \pi (0.30)^2$$

$$\text{Area} = 0.283 \text{ m}^2 \approx 0.28 \text{ m}^2$$

Alternatively: Area = $4 \times \pi r_{mirror}^2 = 4 \times \pi \times 0.15^2$

c) Dependence on distance $d$: Surprising result: The area of illumination is independent of $d$! Why?

- The angle of incidence always equals the angle of reflection
- This is true regardless of the mirror-screen distance
- The geometry scales proportionally with $d$
- The ratio of illuminated radius to mirror radius remains constant at $2:1$
d) Intensity dependence on distance:

While the area stays constant, the intensity (brightness) varies with $d$.

At small distances ($d$ is small):
- Light spreads over a smaller area before hitting the screen
- Intensity is higher, especially at the center
- The illumination is more concentrated
At large distances ($d$ is large):
- Light spreads more before reaching the screen
- Intensity decreases
- The illumination becomes more uniform but dimmer
Summary: The area remains constant at $0.28 \text{ m}^2$ for any distance $d$, but the intensity and brightness distribution vary with the separation.

This problem involves light reflection between angled mirrors and its applications.

Understanding the mirror configuration:

Two plane mirrors form a "V" shape with angle $2\alpha$ between them. - A parallel light beam enters parallel to the axis of symmetry
- The beam reflects back and forth between the mirrors
- Each reflection brings the rays closer to the axis (focusing effect)
a) Determining angles b, c, d, e in terms of $\alpha$:

Understanding the geometry:

When a ray enters at angle $a = \alpha$ to the first mirror:
- Law of reflection: Angle of incidence = Angle of reflection
First reflection (at mirror 1):
- Incident angle: $a = \alpha$
- Reflected angle: $b = \alpha$ (to the normal)
The ray now travels toward mirror 2.

Second reflection (at mirror 2): The angle between the reflected ray and mirror 2 is analyzed.

After working through the geometry:
- Angle $c = 3\alpha$ (incidence at mirror 2)
- Angle $d = 3\alpha$ (reflection from mirror 2)
Third reflection (at mirror 1 again):
- Angle $e = 5\alpha$ (incidence at mirror 1)
Pattern: Each reflection increases the angle by $2\alpha$.
- After reflection 1: $\alpha$
- After reflection 2: $3\alpha$
- After reflection 3: $5\alpha$
- After reflection n: $(2n-1)\alpha$
b) What happens after several reflections:

The angle of incidence increases with each reflection:
- Eventually, the angle approaches or exceeds $90^{\circ}$
- When incidence angle $\geq 90^{\circ}$, the ray can't continue forward
- The ray reflects back out the opening it came from
This creates a "mirror funnel" that concentrates light toward the axis.

c) Number of reflections for $\alpha = 10^{\circ}$: Sequence of angles:
- $a = 10^{\circ}$ (first reflection)
- $c = 30^{\circ}$ (second reflection)
- $e = 50^{\circ}$ (third reflection)
- Fourth reflection: $70^{\circ}$
- Fifth reflection: $90^{\circ}$ (critical angle!)
At $90^{\circ}$, the ray reflects back along its path. Counting reflections:
- Side 1: $10^{\circ}$, $50^{\circ}$ (2 reflections)
- Side 2: $30^{\circ}$, $70^{\circ}$ (2 reflections)
- One more at $90^{\circ}$ where it turns around
Actually, let me recount more carefully:
- Downward path: 4 reflections (alternating mirrors)
- At bottom: 1 reflection at $90^{\circ}$ (turns around)
- Upward path: 4 reflections (same as down)
Total: $4 + 1 + 4 = 9$ reflections d) Why a circular cone works the same:

For rays parallel to the axis of symmetry:
- The incident ray, the normal to the mirror, and the reflected ray all lie in the same plane
- This plane contains the axis of symmetry
- Whether mirrors are flat (polygonal approximation) or perfectly curved:
- The reflection angles are the same
- The ray paths are equivalent
- The calculations remain valid
The cross-section is what matters for rays parallel to the axis!

e) Why the model fails for an ear trumpet: Wave nature of sound:

For light: Wavelength $\lambda_{light} \approx 500$ nm $= 5 \times 10^{-7}$ m
- Trumpet size: $0.5$ m
- Ratio: $\frac{\lambda}{L} \approx 10^{-7}$ (very small!)
Light behaves like rays (geometric optics) because $\lambda \ll$ object size.

For sound: $$\lambda = \frac{v}{f} = \frac{330}{400} = 0.825 \text{ m}$$

- Trumpet length: $0.5$ m
- Wavelength: $0.8$ m
- Ratio: $\frac{\lambda}{L} \approx 1.6$ (comparable!)
When wavelength is comparable to object size, diffraction effects dominate! - Sound doesn't follow simple ray paths
- Wave effects (interference, diffraction) are important
- Geometric optics model breaks down
Summary:
- Angles follow sequence: $\alpha, 3\alpha, 5\alpha, \ldots$
- After enough reflections, ray returns back out
- For $\alpha = 10^{\circ}$: 9 total reflections
- Circular cone equivalent to flat mirrors for on-axis rays
- Sound wave behavior differs due to diffraction (larger $\lambda$)

This problem involves vinyl record physics and wave properties.

Understanding vinyl records:

A vinyl LP (Long Play) record has:
- A single continuous groove spiraling from outer edge to center
- Stylus follows the groove and vibrates to reproduce sound
- Groove width variations encode the audio signal
- Record rotates at constant angular velocity
a) Stylus speed at outer groove:

Given:
- Outer groove radius: $r = 14.6$ cm $= 0.146$ m
- Rotation rate: $33\frac{1}{3}$ rpm $= \frac{100}{3}$ rpm
Convert to angular velocity:

$$\omega = \frac{100}{3} \times \frac{2\pi}{60} = \frac{200\pi}{180} = \frac{10\pi}{9} \text{ rad/s}$$

Linear speed of stylus:

$$v = r\omega = 0.146 \times \frac{10\pi}{9}$$

$$v = 0.51 \text{ m/s}$$

b) Wavelength in groove for 8 kHz: Wave equation: $$v = f\lambda \rightarrow \lambda = \frac{v}{f}$$ Given:
- Frequency: $f = 8.0$ kHz $= 8000$ Hz
$$\lambda = \frac{0.51}{8000} = 6.4 \times 10^{-5} \text{ m}$$

$$\lambda = 64 \text{ micrometers}$$

This is the spatial wavelength along the groove!

c) Wavelength at center of record: Radius at center: $$r_{center} = \frac{2}{5} \times r_{outer} = \frac{2}{5} \times 14.6 = 5.84 \text{ cm}$$ Speed at center: $$v_{center} = \frac{2}{5} \times v_{outer} = \frac{2}{5} \times 0.51 = 0.204 \text{ m/s}$$ Wavelength at center: $$\lambda_{center} = \frac{0.204}{8000} = 2.6 \times 10^{-5} \text{ m}$$

$$\lambda_{center} = 26 \text{ micrometers}$$

Key insight: The wavelength encoded in the groove varies with position (constant frequency requires smaller wavelength at center where stylus moves slower). d) Wavelength in air: Given:
- Speed of sound: $v_{sound} = 330$ m/s
$$\lambda_{air} = \frac{v_{sound}}{f} = \frac{330}{8000}$$

$$\lambda_{air} = 0.041 \text{ m} = 4.1 \text{ cm}$$

Comparison:
- Groove wavelength: $64 \times 10^{-6}$ m
- Air wavelength: $0.041$ m
- Ratio: $\frac{\lambda_{groove}}{\lambda_{air}} \approx 1.6 \times 10^{-3}$
The groove wavelength is about 650 times smaller than the air wavelength! e) Effect of denser gas (CO$_2$): Would the frequency change?

The frequency of the note from the record would NOT change!

Why?

- The frequency is determined by the record groove geometry
- The stylus vibrates at the frequency encoded in the groove
- This is independent of the medium (air, CO$_2$, etc.)
What would change:

- Speed of sound in CO$_2$: Different from air (actually slower)
- Wavelength in CO$_2$: $\lambda = \frac{v_{CO_2}}{f}$ would be different
- The pitch you hear is the frequency, which stays the same
The note itself (frequency) is unchanged, but the sound wave characteristics in the medium are different.

This problem involves intensity calculations and the inverse square law.

Understanding the physical scenario:

We need to find how far away to place a 100 W bulb so it produces the same intensity as the faintest stars in the Milky Way.

Given data:
- Solar intensity at Earth: $I_{\odot} = 1.4$ kW/m$^2$ $= 1400$ W/m$^2$
- Faintest star intensity: $10^{-20} \times$ solar intensity
- Bulb power: $P = 100$ W
Target intensity calculation: Intensity of faintest star: $$I_{target} = 10^{-20} \times I_{\odot}$$

$$I_{target} = 10^{-20} \times 1400 = 1.4 \times 10^{-17} \text{ W/m}^2$$

Inverse square law for light:

For a point source radiating in all directions: $$I = \frac{P}{4\pi r^2}$$

Where:
- $I$ = intensity (W/m$^2$)
- $P$ = power of source (W)
- $r$ = distance from source (m)
- $4\pi r^2$ = surface area of sphere at radius $r$
Solving for distance $r$:

$$I_{target} = \frac{P}{4\pi r^2}$$

$$r^2 = \frac{P}{4\pi I_{target}}$$

$$r^2 = \frac{100}{4\pi \times 1.4 \times 10^{-17}}$$

$$r^2 = \frac{100}{1.76 \times 10^{-16}} = 5.68 \times 10^{17}$$

$$r = \sqrt{5.68 \times 10^{17}}$$

$$r = 7.54 \times 10^{8} \text{ m}$$

$$r \approx 7.5 \times 10^{8} \text{ m}$$

Context:

This distance is about:
- $2.5 \times$ the distance to the Moon ($3.84 \times 10^8$ m)
- About twice the distance to the Sun ($1.5 \times 10^{11}$ m)
A 100 W bulb would need to be placed roughly 750,000 km away to match the intensity of the faintest Milky Way stars!

This problem involves optics, refraction, and viewing through a porthole underwater.

Understanding refraction:

When light passes from water to air:
- Light bends away from the normal (less dense medium)
- Snell's Law: $n_{water}\sin i = n_{air}\sin r$
- Total internal reflection occurs beyond critical angle
Given:
- Water refractive index: $n_{water} = 1.33$
- Porthole radius: $r = 0.1$ m
- Porthole to seabed distance: $d = 4.0$ m
a) Why limited viewing area:

Key concept: The critical angle

Light from outside the porthole can only enter if the angle of incidence in water is less than the critical angle.

Critical angle calculation: $$n_{water}\sin\theta_c = n_{air}\sin 90^{\circ}$$

$$1.33 \times \sin\theta_c = 1 \times 1$$

$$\sin\theta_c = \frac{1}{1.33} = 0.752$$

$$\theta_c = 48.8^{\circ}$$

Only light rays within $48.8^{\circ}$ of the normal can escape the water and reach the observer!

b) Maximum viewing area: Radius of viewing circle:

The maximum viewing radius on the seabed: $$R_{view} = d \tan\theta_c$$

$$R_{view} = 4.0 \times \tan 48.8^{\circ}$$

$$R_{view} = 4.0 \times 1.14 = 4.56 \text{ m}$$

Total radius (including porthole): $$R_{total} = R_{view} + r = 4.56 + 0.1 = 4.66 \text{ m}$$ Area calculation: $$A = \pi R_{total}^2 = \pi(4.66)^2$$

$$A = 68.2 \text{ m}^2 \approx 68 \text{ m}^2$$

The diver can view about $68$ square meters of the seabed through this porthole!

Physical interpretation:

The porthole acts like a window, but refraction limits the field of view. The observer sees a circular patch of the seabed, with the view becoming darker and more distorted toward the edges due to the angle of incidence approaching the critical angle.

This problem involves resonance and sound in pipes.

Understanding pipe resonance:

Sound waves in a pipe form standing waves. For a pipe open at one end:
- Closed end: displacement node (air can't move)
- Open end: displacement antinode (maximum air movement)
Harmonics:

Frequencies: $f_n = \frac{nv}{4L}$ for $n = 1, 3, 5, ...$ (odd harmonics only)

Where $v$ is sound speed and $L$ is pipe length.

Given:
- Two tones from same pipe
- Frequencies: 510 Hz and 680 Hz
Analysis:

The ratio $\frac{680}{510} = \frac{4}{3}$

For consecutive odd harmonics:
- $n$ and $n+2$ give ratio $\frac{n+2}{n}$
- $\frac{n+2}{n} = \frac{4}{3}$ when $n = 6$
Harmonics are 3rd and 5th (or 6th and 8th counting differently).

Fundamental frequency: $$f_1 = \frac{510}{3} = 170 \text{ Hz}$$ Answer: C

$\sin \theta = \frac{v_{\text{sound}}}{v_{\text{plane}}} = \frac{1}{3}$, so $\theta = 19.5^{\circ}$

Minimum distance is $\frac{15}{\sin \theta} = \frac{15}{1/3} = 45$ km