Waves

Given: Mass $M = 2.50 \text{ kg}$. Length $L = 20.0 \text{ m}$. Tension $T = 200 \text{ N}$.

Mass per unit length $\mu = \frac{M}{L} = \frac{2.50}{20.0} = 0.125 \text{ kg m}^{-1}$.

$v = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40 \text{ m s}^{-1}$.

Time taken $t = \frac{L}{v} = \frac{20.0}{40} = \mathbf{0.5 \text{ s}}$.

Height $h = 300 \text{ m}$. $g = 9.8 \text{ m s}^{-2}$. Speed of sound $v_s = 340 \text{ m s}^{-1}$.

1. Time for stone to fall ($t_1$):
$h = \frac{1}{2}gt_1^2 \Rightarrow t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{600}{9.8}} \approx \sqrt{61.22} \approx 7.82 \text{ s}$.

2. Time for sound to travel up ($t_2$):
$t_2 = \frac{h}{v_s} = \frac{300}{340} \approx 0.88 \text{ s}$.

Total Time: $t = t_1 + t_2 = 7.82 + 0.88 = \mathbf{8.7 \text{ s}}$.

Length $L = 12.0 \text{ m}$. Mass $M = 2.10 \text{ kg}$.
$\mu = M/L = 2.10/12.0 = 0.175 \text{ kg m}^{-1}$.
Target Speed $v = 343 \text{ m s}^{-1}$.

$T = (343)^2 \times 0.175 = 117649 \times 0.175 \approx 20588.6 \text{ N}$.
$T \approx \mathbf{2.06 \times 10^4 \text{ N}}$.

Formula: $v = \sqrt{\frac{\gamma P}{\rho}}$. For ideal gas, $PV = nRT \Rightarrow \frac{P}{\rho} = \frac{RT}{M}$. Thus $v = \sqrt{\frac{\gamma RT}{M}}$.

(a) Pressure:
At constant temperature, $P/\rho$ is constant (Boyle’s Law). Thus, $v$ is independent of pressure.

(b) Temperature:
$v \propto \sqrt{T}$. As temperature increases, speed increases.

(c) Humidity:
Moist air has a lower density ($\rho_{moist}$) than dry air ($\rho_{dry}$) because molar mass of water vapor (18) < $N_2/O_2$ (28/32). Since $v \propto \frac{1}{\sqrt{\rho}}$, speed increases with humidity.

A function represents a travelling wave if it is finite and defined for all $x$ and $t$.

(a) $(x – vt)^2$: No. It goes to infinity as $x \to \infty$. Not a finite wave.

(b) $\log[(x + vt)/x_0]$: No. Not defined for negative arguments and goes to infinity.

(c) $1/(x + vt)$: No. It is undefined (singularity) at $x = -vt$.

Frequency $\nu = 1000 \text{ kHz} = 10^6 \text{ Hz}$. This remains constant across media.

(a) Reflected Sound (Air):
$v_{air} = 340 \text{ m s}^{-1}$.
$\lambda_a = \frac{v_{air}}{\nu} = \frac{340}{10^6} = \mathbf{3.4 \times 10^{-4} \text{ m}}$ ($0.34 \text{ mm}$).

(b) Transmitted Sound (Water):
$v_{water} = 1486 \text{ m s}^{-1}$.
$\lambda_w = \frac{v_{water}}{\nu} = \frac{1486}{10^6} = \mathbf{1.49 \times 10^{-3} \text{ m}}$ ($1.49 \text{ mm}$).

Speed $v = 1.7 \text{ km s}^{-1} = 1700 \text{ m s}^{-1}$.
Frequency $\nu = 4.2 \text{ MHz} = 4.2 \times 10^6 \text{ Hz}$.

$\lambda \approx 4.05 \times 10^{-4} \text{ m} = \mathbf{0.4 \text{ mm}}$.

Equation: $y(x, t) = 3.0 \sin (36t + 0.018x + \pi/4)$. ($x, y$ in cm, $t$ in s).

(a) Type & Direction:
Arguments are $(kx + \omega t)$. The positive sign between $x$ and $t$ implies a travelling wave moving in the negative x-direction (Right to Left).
Speed $v = \omega/k = 36/0.018 = 2000 \text{ cm/s} = \mathbf{20 \text{ m/s}}$.

(b) Amplitude & Frequency:
Amplitude $A = \mathbf{3.0 \text{ cm}}$.
$\omega = 36 \Rightarrow 2\pi\nu = 36 \Rightarrow \nu = 36/2\pi \approx \mathbf{5.7 \text{ Hz}}$.

(c) Initial Phase at Origin:
At $x=0, t=0$, phase $\phi = \mathbf{\pi/4}$ rad.

(d) Distance between Crests ($\lambda$):
$k = 0.018 = 2\pi/\lambda \Rightarrow \lambda = 2\pi/0.018 \approx 349 \text{ cm} = \mathbf{3.5 \text{ m}}$.

Eq: $y = 3 \sin(36t + 0.018x + \pi/4)$.

Graphs:
For a fixed $x$, $y(t)$ is a sinusoidal function of time.
At different $x$ values (0, 2, 4 cm), the phase constant $(0.018x + \pi/4)$ changes. The graphs are all sinusoidal with same amplitude (3 cm) and frequency (5.7 Hz), but they are phase-shifted relative to each other.

Differences: Only the phase differs. Amplitude and frequency remain constant for all points in a travelling wave.

$y = 2.0 \cos [2\pi(10t – 0.0080x) + 0.35]$.
Comparing with standard form: $k = 2\pi \times 0.0080 = 0.016\pi \text{ cm}^{-1}$.
Or directly: Phase part dependent on $x$ is $\Phi = \frac{2\pi}{\lambda}x$. Here $\frac{1}{\lambda} = 0.0080$.
$\lambda = 1/0.0080 = 125 \text{ cm} = 1.25 \text{ m}$.
Formula: $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.

(a) $\Delta x = 4$ m = 400 cm:
$\Delta \phi = \frac{2\pi}{125} \times 400 = 6.4\pi \text{ rad}$.

(b) $\Delta x = 0.5$ m = 50 cm:
$\Delta \phi = \frac{2\pi}{125} \times 50 = 0.8\pi \text{ rad}$.

(c) $\Delta x = \lambda/2$:
$\Delta \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \mathbf{\pi \text{ rad}}$.

(d) $\Delta x = 3\lambda/4$:
$\Delta \phi = \frac{2\pi}{\lambda} \times \frac{3\lambda}{4} = \mathbf{1.5\pi \text{ rad}}$.

Equation: $y(x, t) = 0.06 \sin(\frac{2\pi}{3}x) \cos(120\pi t)$.

(a) Type: Since $x$ and $t$ terms are separated (product of functions), it represents a Stationary Wave.

(b) Component Waves:
$2A \sin kx \cos \omega t$ is superposition of $A \sin(kx – \omega t)$ and $A \sin(kx + \omega t)$.
Here $2A = 0.06 \Rightarrow A = 0.03 \text{ m}$.
$k = 2\pi/3 \Rightarrow \lambda = 3 \text{ m}$.
$\omega = 120\pi \Rightarrow \nu = 60 \text{ Hz}$.
Speed $v = \nu \lambda = 60 \times 3 = \mathbf{180 \text{ m/s}}$.
Two waves of amplitude 0.03 m, freq 60 Hz, speed 180 m/s travelling in opposite directions.

(c) Tension:
Mass $m = 3 \times 10^{-2}$ kg. Length $L = 1.5$ m. $\mu = 2 \times 10^{-2}$ kg/m.
$v = \sqrt{T/\mu} \Rightarrow T = v^2 \mu = (180)^2 \times 2 \times 10^{-2} = 32400 \times 0.02 = \mathbf{648 \text{ N}}$.

(i) Oscillations:
In a stationary wave:
(a) Frequency: Same for all points (60 Hz).
(b) Phase: All points in one loop vibrate in phase. Adjacent loops are in opposite phase ($\pi$ difference).
(c) Amplitude: Varies with position $x$. Amplitude function $A(x) = 0.06 \sin(2\pi x/3)$.

(ii) Amplitude at $x = 0.375$ m:
$A = 0.06 \sin(\frac{2\pi}{3} \times 0.375)$
$A = 0.06 \sin(\frac{2\pi}{3} \times \frac{3}{8}) = 0.06 \sin(\frac{\pi}{4})$
$A = 0.06 \times \frac{1}{\sqrt{2}} \approx 0.042 \text{ m} = \mathbf{4.2 \text{ cm}}$.

(a) $y = 2 \cos(3x) \sin(10t)$: Stationary Wave. (Harmonic $x$ and $t$ terms separated).

(b) $y = 2\sqrt{x – vt}$: Travelling Wave. Function of $(x-vt)$. Valid only for $x \ge vt$.

(c) $y = 3 \sin(5x – 0.5t) + 4 \cos(5x – 0.5t)$: Travelling Wave. Superposition of two travelling waves with same phase velocity. Can be reduced to $A \sin(kx – \omega t + \phi)$.

(d) $y = \cos x \sin t + \cos 2x \sin 2t$: Stationary Wave. Superposition of two stationary waves.

Given: $\nu = 45 \text{ Hz}$ (Fundamental).
Mass $m = 3.5 \times 10^{-2} \text{ kg}$. $\mu = 4.0 \times 10^{-2} \text{ kg m}^{-1}$.
Length $L = m/\mu = 3.5/4.0 = 0.875 \text{ m}$.

(a) Wave Speed:
For fundamental mode, $L = \lambda/2 \Rightarrow \lambda = 2L = 1.75 \text{ m}$.
$v = \nu \lambda = 45 \times 1.75 = \mathbf{78.75 \text{ m/s}}$.

(b) Tension:
$v = \sqrt{T/\mu} \Rightarrow T = v^2 \mu = (78.75)^2 \times 4.0 \times 10^{-2}$.
$T \approx \mathbf{248 \text{ N}}$.

Frequency $\nu = 340 \text{ Hz}$.
Resonance lengths $L_1 = 25.5 \text{ cm}$, $L_2 = 79.3 \text{ cm}$.
For closed pipe, $L_1 + e = \lambda/4$ and $L_2 + e = 3\lambda/4$.
Subtracting: $L_2 – L_1 = \lambda/2$.
$\lambda = 2(79.3 – 25.5) = 2(53.8) = 107.6 \text{ cm} = 1.076 \text{ m}$.
Speed $v = \nu \lambda = 340 \times 1.076 = \mathbf{365.8 \text{ m/s}}$.

Rod clamped at middle $\rightarrow$ Node at center, Antinodes at ends.
Length $L = 100 \text{ cm} = 1 \text{ m}$.
Fundamental Mode: $L = \lambda/2$ (Distance between two antinodes).
$\lambda = 2L = 2 \text{ m}$.
Frequency $\nu = 2.53 \text{ kHz} = 2530 \text{ Hz}$.
Speed $v = \nu \lambda = 2530 \times 2 = \mathbf{5060 \text{ m/s}} = \mathbf{5.06 \text{ km/s}}$.

Length $L = 20 \text{ cm} = 0.2 \text{ m}$. Source $\nu = 430 \text{ Hz}$. Speed $v = 340 \text{ m/s}$.
Fundamental freq (Closed): $\nu_0 = v/4L = 340/(4 \times 0.2) = 340/0.8 = 425 \text{ Hz}$.
The source frequency (430 Hz) is very close to the fundamental (425 Hz). It will excite the 1st Harmonic (Fundamental Mode) resonantly.

Open Pipe:
Fundamental $\nu’_0 = v/2L = 340/0.4 = 850 \text{ Hz}$.
Harmonics: 850, 1700…
430 Hz is not close to any harmonic of the open pipe. No resonance.

Frequency $f_A = 324 \text{ Hz}$. Beat freq $b = 6 \text{ Hz}$.
Possibilities for $f_B$: $324 \pm 6$, i.e., $330 \text{ Hz}$ or $318 \text{ Hz}$.
Condition: Tension in A reduced $\rightarrow f_A$ decreases.
New beat frequency $b’ = 3 \text{ Hz}$.

Case 1: $f_B = 330$ Hz.
If $f_A$ drops (say to 320), beats $|330-320| = 10$. Increases. (Incorrect).

Case 2: $f_B = 318$ Hz.
If $f_A$ drops (say to 321), beats $|318-321| = 3$. Decreases. (Correct).

So, frequency of B is 318 Hz.

(a) Displacement Node = Pressure Antinode:
At a displacement node, particles are not moving, but particles from both sides are converging/diverging maximally, causing maximum compression/rarefaction (pressure variation).

(b) Bat Echolocation:
Bats emit ultrasonic pulses and detect the reflected echoes. The time delay gives distance; the intensity and frequency shift (Doppler) give nature and speed of obstacles.

(c) Timbre/Quality:
Violin and sitar notes have different overtone structures (harmonics). Even with the same fundamental frequency, the mix of higher harmonics differs, giving distinct quality.

(d) Waves in Solids vs Gases:
Solids have shear modulus (rigidity) to support transverse waves. Gases lack rigidity and can only sustain compression (longitudinal waves).

(e) Pulse Distortion:
In a dispersive medium, wave speed depends on frequency. A pulse is a superposition of many frequencies. Different components travel at different speeds, spreading out the pulse shape.