Waves

Given:
- Mass of string, $m = 2.50$ kg
- Tension, $T = 200$ N
- Length, $L = 20.0$ m

Formula used:
Speed of transverse wave on a string: $$v = \sqrt{\frac{T}{\mu}}$$
where $\mu$ = linear mass density = $\dfrac{m}{L}$

Step 1: Find linear mass density
$$\mu = \frac{m}{L} = \frac{2.50}{20.0} = 0.125 \text{ kg m}^{-1}$$

Step 2: Find speed of wave
$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40 \text{ m s}^{-1}$$

Step 3: Find time to travel length $L$
$$t = \frac{L}{v} = \frac{20.0}{40} = 0.50 \text{ s}$$

Answer: The disturbance takes 0.50 s to reach the other end.

Given:
- Height of tower, $h = 300$ m
- Speed of sound, $v_s = 340$ m s$^{-1}$
- $g = 9.8$ m s$^{-2}$

Step 1: Time for stone to fall to the base ($t_1$)
Using $h = \dfrac{1}{2}g t_1^2$:
$$t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 300}{9.8}} = \sqrt{61.22} \approx 7.82 \text{ s}$$

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

Step 3: Total time
$$t = t_1 + t_2 = 7.82 + 0.88 = 8.70 \text{ s}$$

Answer: The splash is heard at the top after approximately 8.70 s.

Given:
- Length of wire, $L = 12.0$ m
- Mass of wire, $m = 2.10$ kg
- Required speed, $v = 343$ m s$^{-1}$

Step 1: Find linear mass density
$$\mu = \frac{m}{L} = \frac{2.10}{12.0} = 0.175 \text{ kg m}^{-1}$$

Step 2: Use formula $v = \sqrt{T/\mu}$ and solve for $T$
$$v^2 = \frac{T}{\mu} \implies T = \mu v^2$$
$$T = 0.175 \times (343)^2 = 0.175 \times 117649$$
$$T \approx 2.06 \times 10^4 \text{ N}$$

Answer: The required tension is approximately $T \approx 2.06 \times 10^4$ N.

Formula: $v = \sqrt{\dfrac{\gamma P}{\rho}}$

(a) Independent of pressure:
For an ideal gas: $PV = nRT \Rightarrow P = \dfrac{\rho RT}{M}$, where $M$ is the molar mass.

Substituting:
$$v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M}}$$

At constant temperature, when pressure $P$ increases, density $\rho$ also increases proportionally (since $\rho \propto P$ at constant $T$). Therefore the ratio $P/\rho$ remains constant. Hence speed of sound is independent of pressure.

(b) Increases with temperature:
From the expression $v = \sqrt{\dfrac{\gamma RT}{M}}$, we see that $v \propto \sqrt{T}$.

As temperature $T$ increases, $v$ increases. Hence speed of sound increases with temperature.

(c) Increases with humidity:
Moist air contains water vapour. The molar mass of water ($M_{\text{H}_2\text{O}} = 18$ g/mol) is less than the effective molar mass of dry air ($M_{\text{air}} \approx 29$ g/mol).

Since $v = \sqrt{\dfrac{\gamma RT}{M}}$, a smaller molar mass $M$ gives a larger $v$. Humid air has a lower effective molar mass (and hence lower density) than dry air at the same temperature and pressure. Therefore speed of sound increases with humidity.

Concept: For a function to represent a physically valid travelling wave, it must:
1. Be finite and well-defined for all $x$ and $t$.
2. Be single-valued.
3. The converse is not necessarily true — a function of $(x \pm vt)$ need not represent a travelling wave unless it satisfies the above physical conditions.

(a) $y = (x - vt)^2$:
This is a function of $(x - vt)$, so it satisfies the mathematical form. However, as $x \to \infty$ or $t \to \infty$, $y \to \infty$. A wave must have a finite displacement. This does not represent a travelling wave.

(b) $y = \log\left[\dfrac{x + vt}{x_0}\right]$:
This is a function of $(x + vt)$. However, $\log$ is not defined for negative arguments, and as $(x + vt) \to 0$, $y \to -\infty$, and as $(x + vt) \to \infty$, $y \to \infty$. The displacement is not bounded. This does not represent a travelling wave.

(c) $y = \dfrac{1}{x + vt}$:
This is a function of $(x + vt)$. However, when $x + vt = 0$, $y \to \infty$, which is physically impossible. This does not represent a travelling wave.

Conclusion: None of the three functions represent a valid travelling wave, even though (a), (b), and (c) are functions of $(x \pm vt)$. The converse is not true in general.

Given:
- Frequency, $\nu = 1000$ kHz $= 10^6$ Hz
- Speed of sound in air, $v_{\text{air}} = 340$ m s$^{-1}$
- Speed of sound in water, $v_{\text{water}} = 1486$ m s$^{-1}$

Key concept: When a wave crosses a boundary, its frequency remains unchanged. Only the speed (and hence wavelength) changes.

(a) Wavelength of reflected sound (in air):
The reflected wave travels back in air, so:
$$\lambda_{\text{reflected}} = \frac{v_{\text{air}}}{\nu} = \frac{340}{10^6} = 3.4 \times 10^{-4} \text{ m}$$

(b) Wavelength of transmitted sound (in water):
The transmitted wave travels in water:
$$\lambda_{\text{transmitted}} = \frac{v_{\text{water}}}{\nu} = \frac{1486}{10^6} = 1.486 \times 10^{-3} \text{ m}$$

Answer:
- Wavelength of reflected sound: $\lambda = 3.4 \times 10^{-4}$ m
- Wavelength of transmitted sound: $\lambda = 1.486 \times 10^{-3}$ m

Given:
- Speed of sound in tissue, $v = 1.7$ km s$^{-1}$ $= 1.7 \times 10^3$ m s$^{-1}$
- Frequency, $\nu = 4.2$ MHz $= 4.2 \times 10^6$ Hz

Formula: $\lambda = \dfrac{v}{\nu}$

$$\lambda = \frac{1.7 \times 10^3}{4.2 \times 10^6} = \frac{1.7}{4.2} \times 10^{-3} \approx 0.405 \times 10^{-3} \text{ m}$$

$$\lambda \approx 4.05 \times 10^{-4} \text{ m}$$

Answer: The wavelength of sound in the tissue is approximately $4.05 \times 10^{-4}$ m (about 0.4 mm).

Given wave: $y(x,t) = 3.0\sin(36t + 0.018x + \pi/4)$ cm

Comparing with standard form $y = a\sin(\omega t + kx + \phi)$:
- $a = 3.0$ cm, $\omega = 36$ rad s$^{-1}$, $k = 0.018$ rad cm$^{-1}$, $\phi = \pi/4$

(a) Nature and direction of wave:
Since the equation contains $(\omega t + kx)$, i.e., $x$ and $t$ appear as $(x + vt)$, this is a travelling wave moving in the negative x-direction (from right to left).

Speed of propagation:
$$v = \frac{\omega}{k} = \frac{36}{0.018} = 2000 \text{ cm s}^{-1} = 20 \text{ m s}^{-1}$$

(b) Amplitude and frequency:
$$\text{Amplitude} = a = 3.0 \text{ cm}$$
$$\text{Frequency} = \nu = \frac{\omega}{2\pi} = \frac{36}{2\pi} \approx 5.73 \text{ Hz}$$

(c) Initial phase at origin:
At $x = 0$, $t = 0$:
$$\text{Phase} = \phi = \frac{\pi}{4}$$

(d) Least distance between two successive crests (wavelength):
$$\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.018} \approx 349 \text{ cm} \approx 3.49 \text{ m}$$

Answer Summary:
- Travelling wave, moving in negative x-direction at 20 m s⁻¹
- Amplitude = 3.0 cm, Frequency ≈ 5.73 Hz
- Initial phase = $\pi/4$
- Wavelength (distance between crests) ≈ 3.49 m

Wave equation: $y(x,t) = 3.0\sin(36t + 0.018x + \pi/4)$ cm

At $x = 0$:
$$y(0,t) = 3.0\sin\left(36t + \frac{\pi}{4}\right)$$

At $x = 2$ cm:
$$y(2,t) = 3.0\sin\left(36t + 0.018 \times 2 + \frac{\pi}{4}\right) = 3.0\sin\left(36t + 0.036 + \frac{\pi}{4}\right)$$

At $x = 4$ cm:
$$y(4,t) = 3.0\sin\left(36t + 0.018 \times 4 + \frac{\pi}{4}\right) = 3.0\sin\left(36t + 0.072 + \frac{\pi}{4}\right)$$

Shape of graphs: All three graphs are sinusoidal (sine curves) with the same amplitude (3.0 cm) and the same frequency ($\nu \approx 5.73$ Hz), but they are shifted in phase with respect to each other.

Difference between oscillatory motions at different points:
- Amplitude: Same (3.0 cm) at all points — no difference.
- Frequency: Same ($\approx 5.73$ Hz) at all points — no difference.
- Phase: Different at different points. The phase increases with $x$ (since the wave moves in the $-x$ direction, points at larger $x$ are ahead in phase).

Conclusion: In a travelling wave, all particles oscillate with the same amplitude and frequency, but with different phases.

Given wave: $y(x,t) = 2.0\cos 2\pi(10t - 0.0080x + 0.35)$

Rewriting: $y = 2.0\cos(20\pi t - 0.016\pi x + 0.70\pi)$

Comparing with $y = a\cos(\omega t - kx + \phi)$:
$$k = 0.016\pi \text{ rad cm}^{-1} = 0.016\pi \times 100 \text{ rad m}^{-1} = 1.6\pi \text{ rad m}^{-1}$$

Wavelength:
$$\lambda = \frac{2\pi}{k} = \frac{2\pi}{1.6\pi} = 1.25 \text{ m}$$

Phase difference formula:
$$\Delta\phi = k \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \Delta x$$

(a) $\Delta x = 4$ m:
$$\Delta\phi = \frac{2\pi}{1.25} \times 4 = \frac{8\pi}{1.25} = 6.4\pi \text{ rad} \approx 20.1 \text{ rad}$$

(b) $\Delta x = 0.5$ m:
$$\Delta\phi = \frac{2\pi}{1.25} \times 0.5 = \frac{\pi}{1.25} = 0.8\pi \text{ rad} \approx 2.51 \text{ rad}$$

(c) $\Delta x = \lambda/2 = 0.625$ m:
$$\Delta\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} = \pi \text{ rad}$$

(d) $\Delta x = 3\lambda/4 = 0.9375$ m:
$$\Delta\phi = \frac{2\pi}{\lambda} \times \frac{3\lambda}{4} = \frac{3\pi}{2} \text{ rad}$$

Answers:
- (a) $\Delta\phi = 6.4\pi$ rad
- (b) $\Delta\phi = 0.8\pi$ rad
- (c) $\Delta\phi = \pi$ rad
- (d) $\Delta\phi = 3\pi/2$ rad

Given: $y(x,t) = 0.06\sin\left(\dfrac{2\pi}{3}x\right)\cos(120\pi t)$
- Length $L = 1.5$ m, Mass $m = 3.0 \times 10^{-2}$ kg

(a) Nature of wave:
The equation is of the form $y = (2a\sin kx)\cos\omega t$, which is the standard form of a stationary (standing) wave. It has nodes and antinodes at fixed positions.

(b) Superposition interpretation:
Using the identity: $\sin A \cos B = \dfrac{1}{2}[\sin(A+B) + \sin(A-B)]$:
$$y = 0.03\sin\left(\frac{2\pi}{3}x + 120\pi t\right) + 0.03\sin\left(\frac{2\pi}{3}x - 120\pi t\right)$$

This represents two waves of amplitude 0.03 m travelling in opposite directions.

From $k = \dfrac{2\pi}{3}$ rad m$^{-1}$ and $\omega = 120\pi$ rad s$^{-1}$:

$$\lambda = \frac{2\pi}{k} = \frac{2\pi}{2\pi/3} = 3 \text{ m}$$

$$\nu = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60 \text{ Hz}$$

$$v = \frac{\omega}{k} = \frac{120\pi}{2\pi/3} = 180 \text{ m s}^{-1}$$

(c) Tension in the string:
Linear mass density:
$$\mu = \frac{m}{L} = \frac{3.0 \times 10^{-2}}{1.5} = 2.0 \times 10^{-2} \text{ kg m}^{-1}$$

Using $v = \sqrt{T/\mu}$:
$$T = \mu v^2 = 2.0 \times 10^{-2} \times (180)^2 = 2.0 \times 10^{-2} \times 32400$$
$$T = 648 \text{ N}$$

Answer: Tension in the string = 648 N

Wave equation: $y(x,t) = 0.06\sin\left(\dfrac{2\pi}{3}x\right)\cos(120\pi t)$

(i)(a) Frequency:
Yes, all points oscillate with the same frequency. The time-dependent part $\cos(120\pi t)$ is the same for all $x$, giving $\nu = 60$ Hz for every point.

(i)(b) Phase:
All points between two consecutive nodes oscillate in phase with each other. Points on opposite sides of a node are in opposite phase (phase difference of $\pi$). So not all points have the same phase.

(i)(c) Amplitude:
No. The amplitude of oscillation at position $x$ is $A(x) = 0.06\sin\left(\dfrac{2\pi}{3}x\right)$, which varies with $x$. Points at nodes have zero amplitude; points at antinodes have maximum amplitude (0.06 m).

(ii) Amplitude at $x = 0.375$ m:
$$A = 0.06\sin\left(\frac{2\pi}{3} \times 0.375\right) = 0.06\sin\left(\frac{2\pi \times 0.375}{3}\right)$$
$$= 0.06\sin\left(\frac{0.75\pi}{3}\right) = 0.06\sin(0.25\pi) = 0.06\sin\left(\frac{\pi}{4}\right)$$
$$= 0.06 \times \frac{1}{\sqrt{2}} = 0.06 \times 0.7071 \approx 0.0424 \text{ m}$$

Answer: The amplitude at $x = 0.375$ m is approximately 0.042 m (4.2 cm).

(a) $y = 2\cos(3x)\sin(10t)$:
This is a product of a function of $x$ alone and a function of $t$ alone. This is the standard form of a stationary (standing) wave.

→ (ii) Stationary wave

(b) $y = 2\sqrt{x - vt}$:
This is a function of $(x - vt)$, so it satisfies the mathematical form of a travelling wave. However, $\sqrt{x - vt}$ is not defined for x < vt (negative argument under square root), and it is not bounded (as $x - vt \to \infty$, $y \to \infty$). It does not represent a physically valid wave.

→ (iii) None at all

(c) $y = 3\sin(5x - 0.5t) + 4\cos(5x - 0.5t)$:
Both terms are functions of $(5x - 0.5t)$, i.e., of the form $f(x - vt)$. They can be combined:
$$y = R\sin(5x - 0.5t + \delta)$$
where $R = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5$ and $\tan\delta = 4/3$.
This is a single sinusoidal wave travelling in the positive $x$-direction.

→ (i) Travelling wave

(d) $y = \cos x \sin t + \cos 2x \sin 2t$:
This is a sum of two stationary wave terms. Each term is a product of a function of $x$ and a function of $t$. However, the two terms have different wavelengths and frequencies, so the superposition does not form a simple stationary wave pattern. It is a superposition of two standing waves.

→ (ii) Stationary wave (superposition of two standing waves)

Given:
- Fundamental frequency, $\nu_1 = 45$ Hz
- Mass of wire, $m = 3.5 \times 10^{-2}$ kg
- Linear mass density, $\mu = 4.0 \times 10^{-2}$ kg m$^{-1}$

Step 1: Find length of wire
$$L = \frac{m}{\mu} = \frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}} = 0.875 \text{ m}$$

(a) Speed of transverse wave:
For fundamental mode (first harmonic) of a string fixed at both ends:
$$\nu_1 = \frac{v}{2L}$$
$$v = 2L\nu_1 = 2 \times 0.875 \times 45 = 78.75 \text{ m s}^{-1}$$

$$\boxed{v \approx 78.75 \text{ m s}^{-1}}$$

(b) Tension in the string:
Using $v = \sqrt{T/\mu}$:
$$T = \mu v^2 = 4.0 \times 10^{-2} \times (78.75)^2$$
$$T = 4.0 \times 10^{-2} \times 6201.56 \approx 248 \text{ N}$$

$$\boxed{T \approx 248 \text{ N}}$$

Given:
- Frequency of tuning fork, $\nu = 340$ Hz
- First resonance length, $L_1 = 25.5$ cm $= 0.255$ m
- Second resonance length, $L_2 = 79.3$ cm $= 0.793$ m

Concept: For a tube closed at one end (piston end) and open at the other, resonance occurs at odd multiples of $\lambda/4$. Consecutive resonance lengths differ by $\lambda/2$.

Step 1: Find wavelength
$$L_2 - L_1 = \frac{\lambda}{2}$$
$$\lambda = 2(L_2 - L_1) = 2(0.793 - 0.255) = 2 \times 0.538 = 1.076 \text{ m}$$

Step 2: Find speed of sound
$$v = \nu\lambda = 340 \times 1.076 = 365.84 \text{ m s}^{-1}$$

$$\boxed{v \approx 366 \text{ m s}^{-1}}$$

Answer: The speed of sound in air at the temperature of the experiment is approximately 366 m s⁻¹.

Given:
- Length of rod, $L = 100$ cm $= 1.0$ m
- Fundamental frequency, $\nu = 2.53$ kHz $= 2530$ Hz

Concept: When a rod is clamped at its middle, the middle point is a node and both free ends are antinodes. For the fundamental mode:
$$L = \frac{\lambda}{2} \implies \lambda = 2L$$

Step 1: Find wavelength
$$\lambda = 2L = 2 \times 1.0 = 2.0 \text{ m}$$

Step 2: Find speed of sound
$$v = \nu\lambda = 2530 \times 2.0 = 5060 \text{ m s}^{-1}$$

$$\boxed{v = 5060 \text{ m s}^{-1} \approx 5.06 \text{ km s}^{-1}}$$

Answer: The speed of sound in steel is approximately 5060 m s⁻¹.

Given:
- Length of pipe, $L = 20$ cm $= 0.20$ m
- Frequency of source, $\nu = 430$ Hz
- Speed of sound, $v = 340$ m s$^{-1}$

Case 1: Pipe closed at one end
Frequencies of normal modes:
$$\nu_n = (2n-1)\frac{v}{4L}, \quad n = 1, 2, 3, \ldots$$
$$\frac{v}{4L} = \frac{340}{4 \times 0.20} = \frac{340}{0.80} = 425 \text{ Hz}$$

For $n = 1$: $\nu_1 = 425$ Hz
For $n = 2$: $\nu_2 = 3 \times 425 = 1275$ Hz

The source frequency 430 Hz does not exactly match. However, checking:
$$n = \frac{\nu}{v/(4L)} = \frac{430}{425} \approx 1.01$$

This is very close to $n = 1$ (fundamental mode). In practice, with slight end corrections, the first harmonic (fundamental mode) of the closed pipe is resonantly excited.

Case 2: Pipe open at both ends
Frequencies of normal modes:
$$\nu_n = n\frac{v}{2L}, \quad n = 1, 2, 3, \ldots$$
$$\frac{v}{2L} = \frac{340}{2 \times 0.20} = 850 \text{ Hz}$$

For $n = 1$: $\nu_1 = 850$ Hz
For $n = 2$: $\nu_2 = 1700$ Hz

None of these match 430 Hz. The 430 Hz source will NOT be in resonance with the open pipe.

Answer:
- Closed pipe: First harmonic (fundamental mode) is resonantly excited.
- Open pipe: No resonance with the 430 Hz source.

Given:
- Original beat frequency = 6 Hz
- Original frequency of A, $\nu_A = 324$ Hz
- After reducing tension in A, beat frequency = 3 Hz

Step 1: Find possible frequency of B
Beat frequency $= |\nu_A - \nu_B| = 6$ Hz
$$\nu_B = 324 \pm 6 \text{ Hz}$$
So $\nu_B = 330$ Hz or $\nu_B = 318$ Hz.

Step 2: Use the effect of reducing tension
When tension in string A is reduced, its frequency decreases (since $\nu \propto \sqrt{T}$).

- If $\nu_B = 330$ Hz: Reducing $\nu_A$ below 324 Hz would increase the difference $|\nu_A - \nu_B|$, so beat frequency would increase (not decrease to 3 Hz). ✗

- If $\nu_B = 318$ Hz: Reducing $\nu_A$ below 324 Hz brings it closer to 318 Hz, so the difference $|\nu_A - \nu_B|$ decreases. Beat frequency reduces to 3 Hz. ✓

Answer: The frequency of string B is $\nu_B =$ 330 Hz.

*Wait — let me re-examine:* If $\nu_B = 330$ Hz and $\nu_A$ decreases, $\nu_A$ moves away from 330, beats increase. If $\nu_B = 318$ Hz and $\nu_A$ decreases from 324, it moves toward 318, beats decrease from 6 to 3. This is consistent.

$$\boxed{\nu_B = 318 \text{ Hz}}$$

(a) Displacement node is a pressure antinode and vice versa:

In a sound wave, displacement and pressure variations are related. At a displacement node, the particles do not move but the layers on either side are alternately compressed and rarefied — this results in maximum pressure variation (pressure antinode). Conversely, at a displacement antinode, particles have maximum displacement but the pressure variation is minimum (pressure node), because the medium is neither compressed nor rarefied at that point. Mathematically, pressure variation is proportional to $-\partial y/\partial x$; a node in $y$ corresponds to maximum $\partial y/\partial x$, hence maximum pressure variation.

(b) Bats and echolocation:

Bats emit high-frequency ultrasonic sound pulses (up to ~100 kHz) and listen to the echoes reflected from obstacles. By analysing the time delay between emission and reception of the echo, they determine distance (since $d = v \cdot t/2$). The direction is determined by comparing the intensity and time of arrival at the two ears. The nature and size of the obstacle are determined from the intensity, frequency shift (Doppler effect), and pattern of the reflected pulse. This biological sonar system allows bats to navigate and hunt in complete darkness.

(c) Same frequency but different quality (timbre):

A violin and a sitar string vibrating at the same fundamental frequency produce notes of the same pitch. However, the two instruments produce different overtones (harmonics) with different relative amplitudes. The quality or timbre of a note depends on the number and relative intensities of the harmonics present. Since the two instruments have different shapes, sizes, and materials, their harmonic content differs, allowing us to distinguish between them even at the same frequency.

(d) Longitudinal and transverse waves in solids vs. gases:

Transverse waves require a restoring force for shear deformation, i.e., they need a shear modulus (modulus of rigidity). Solids possess both bulk modulus and shear modulus, so they can support both longitudinal and transverse waves. Gases and liquids have no shear modulus (they cannot sustain shear stress — they flow). Therefore, only longitudinal (compressional) waves, which require bulk modulus, can propagate in gases.

(e) Distortion of pulse in a dispersive medium:

A pulse is not a single frequency wave; it is a superposition of waves of many different frequencies (Fourier components). In a dispersive medium, the speed of propagation depends on frequency ($v = v(\nu)$). Different frequency components of the pulse travel at different speeds. As the pulse propagates, these components get out of phase with each other, causing the shape of the pulse to change (distort). In a non-dispersive medium, all components travel at the same speed and the pulse shape is preserved.