Chapter 14: Waves (Oscillations and Waves)

$\mu =\frac{2.50}{20.0}=0.125{\text{kg m}}^{-1}$; $v=\sqrt{\frac{200}{0.125}}=40{\text{m s}}^{-1}$; $t=\frac{20.0}{40}$.

0.5 s

Fall: $t_1=\sqrt{\frac{2\times 300}{9.8}}=7.82$ s. Sound up: $t_2=\frac{300}{340}=0.88$ s.

After 8.7 s

$\mu =\frac{2.10}{12.0}=0.175{\text{kg m}}^{-1}$; $T=\mu v^2=0.175\times (343{)}^2$.

2.06 × 10⁴ N

For an ideal gas $P=\frac{\rho RT}{M}$, so $v=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{\gamma RT}{M}}$.

(a) At constant temperature, P/ρ is constant (doubling P doubles ρ), so v does not depend on pressure.

(b) $v\propto \sqrt{T}$, so v increases with temperature.

(c) Water vapour (M = 18) is lighter than N₂ (28) and O₂ (32). Humid air has a smaller average molar mass (smaller density), so v is larger.

v = √(γRT/M): (a) no P in it (b) v ∝ √T (c) moist air has lower M.

The converse is not true. Besides having the form $f(x\pm vt)$, a physical wave must be finite everywhere and at all times.

(a) $(x-vt{)}^2$ grows without limit as x increases: not acceptable.

(b) $\log \frac{x+vt}{x_0}$ is undefined for $x+vt\le 0$ and unbounded: not acceptable.

(c) $\frac{1}{x+vt}$ falls to zero far away but blows up at the single point $x=-vt$. The NCERT answer accepts (c) as the only possible one; strictly, the infinity at $x=-vt$ also makes it unphysical.

The converse is not true; (a) and (b) cannot be travelling waves; NCERT accepts only (c).

Frequency does not change on reflection or refraction; $\nu =10^6$ Hz.

(a) Reflected (in air): $\lambda =\frac{340}{10^6}=3.4\times 10^{-4}$ m. (b) Transmitted (in water): $\lambda =\frac{1486}{10^6}=1.49\times 10^{-3}$ m.

(a) 3.4 × 10⁻⁴ m (b) 1.49 × 10⁻³ m

$\lambda =\frac{v}{\nu }=\frac{1.7\times 10^3}{4.2\times 10^6}$

4.1 × 10⁻⁴ m

Compare with $y=a\sin (\omega t+kx+\varphi )$: a = 3.0 cm, ω = 36 rad s⁻¹, k = 0.018 cm⁻¹, φ = π/4.

(a) Travelling wave; $(\omega t+kx)$ means it moves towards −x (right to left). $v=\frac{\omega }{k}=\frac{36}{0.018}=2000$ cm s⁻¹ $=20$ m s⁻¹.

(b) Amplitude 3.0 cm; $\nu =\frac{36}{2\pi }=5.7$ Hz.

(c) π/4.

(d) $\lambda =\frac{2\pi }{k}=\frac{2\pi }{0.018}=349$ cm.

(a) Travelling, 20 m s⁻¹, right to left (b) 3.0 cm, 5.7 Hz (c) π/4 (d) 3.5 m

$y(0,t)=3\sin \left(36t+\frac{\pi }{4}\right)$, $y(2,t)=3\sin \left(36t+0.036+\frac{\pi }{4}\right)$, $y(4,t)=3\sin \left(36t+0.072+\frac{\pi }{4}\right)$.

All three graphs are sine curves with amplitude 3 cm and period $\frac{2\pi }{36}=0.17$ s, shifted slightly in time.

All are sinusoidal with the same amplitude and frequency; they differ only in initial phase.

$k=2\pi \times 0.0080=0.016\pi$ rad cm⁻¹, so $\lambda =\frac{1}{0.0080}=125$ cm. Phase difference $=k\Delta x$.

(a) $0.016\pi \times 400=6.4\pi$ rad (b) $0.016\pi \times 50=0.8\pi$ rad (c) $\frac{2\pi }{\lambda }\cdot \frac{\lambda }{2}=\pi$ rad (d) $\frac{2\pi }{\lambda }\cdot \frac{3\lambda }{4}=\frac{3\pi }{2}$ rad

(a) 6.4π rad (b) 0.8π rad (c) π rad (d) 3π/2 rad

(a) It is a product of a function of x and a function of t: a stationary wave.

(b) $2\sin A\cos B=\sin (A+B)+\sin (A-B)$, so it is the sum of $0.03\sin \left(\frac{2\pi x}{3}+120\pi t\right)$ and $0.03\sin \left(\frac{2\pi x}{3}-120\pi t\right)$, moving in opposite directions.

$k=\frac{2\pi }{3}\Rightarrow \lambda =3$ m; $\omega =120\pi \Rightarrow \nu =60$ Hz; $v=\nu \lambda =180$ m s⁻¹.

(c) $\mu =\frac{3.0\times 10^{-2}}{1.5}=0.02$ kg m⁻¹; $T=\mu v^2=0.02\times 180^2=648$ N.

(a) Stationary (b) λ = 3 m, ν = 60 Hz, v = 180 m s⁻¹ each (c) 648 N

(i) (a) Yes, all points (except nodes, which are at rest) oscillate at 60 Hz. (b) The string (1.5 m = λ/2) forms a single loop, so all its points move in the same phase. (c) No: the amplitude $0.06\sin \frac{2\pi x}{3}$ depends on position.

(ii) $A=0.06\sin \left(\frac{2\pi \times 0.375}{3}\right)=0.06\sin \frac{\pi }{4}=0.042$ m.

(i) Same frequency and phase, different amplitudes (ii) 0.042 m

(a) Product of a function of x and a function of t: stationary wave.

(b) Not defined for x < vt and grows without limit: not acceptable for any wave.

(c) Both terms have the argument $(5x-0.5t)$; together they form $5\sin (5x-0.5t+\varphi )$: travelling harmonic wave.

(d) Sum of two stationary waves (each a product form).

(a) Stationary (b) none (c) travelling (d) superposition of two stationary waves

Length $L=\frac{3.5\times 10^{-2}}{4.0\times 10^{-2}}=0.875$ m. Fundamental: $\lambda =2L=1.75$ m.

(a) $v=\nu \lambda =45\times 1.75=78.75{\text{m s}}^{-1}$ (b) $T=\mu v^2=0.04\times (78.75{)}^2=248$ N

(a) 79 m s⁻¹ (b) 248 N

The tube is closed at the piston end, so resonance occurs at $L=\frac{\lambda }{4},\frac{3\lambda }{4},\dots$ (odd harmonics; $79.3\approx 3\times 25.5$ confirms this).

From the first resonance: $\frac{\lambda }{4}=0.255$ m $\Rightarrow \lambda =1.02$ m, $v=\nu \lambda =340\times 1.02$.

About 347 m s⁻¹

The clamped middle is a node and the free ends are antinodes, so in the fundamental mode $L=\frac{\lambda }{2}\Rightarrow \lambda =2.0$ m.

$v=\nu \lambda =2.53\times 10^3\times 2.0$

5.06 km s⁻¹

Closed pipe: ${\nu }_1=\frac{v}{4L}=\frac{340}{0.80}=425$ Hz ≈ 430 Hz: the fundamental (first harmonic). (Its other modes, 1275 Hz, … are far away.)

Open at both ends: ${\nu }_1=\frac{v}{2L}=850$ Hz, and the higher modes are 1700 Hz, …; none is near 430 Hz.

The fundamental (first harmonic); no resonance if both ends are open.

${\nu }_B=324\pm 6=330$ or 318 Hz. Reducing the tension lowers ${\nu }_A$ (as $\nu \propto \sqrt{T}$).

If ${\nu }_B=330$, lowering ${\nu }_A$ would increase the beats. If ${\nu }_B=318$, lowering ${\nu }_A$ towards 318 reduces them (to 3 Hz). So ${\nu }_B=318$ Hz.

318 Hz

(a) At a displacement node, the air on both sides moves towards it (or away from it) together, so it is compressed and rarefied the most: the pressure change is greatest there. Where the displacement is largest (antinode), neighbouring layers move together and the pressure hardly changes.

(b) Bats emit ultrasonic pulses and listen to the echoes. The delay gives distance, the direction of the echo gives direction, and the strength and quality of the reflected sound tell the nature and size of the obstacle.

(c) The two instruments produce the same fundamental but different overtones (harmonics) with different relative strengths. This gives each note its own quality (timbre).

(d) Transverse waves need a shear (rigidity) restoring force. Solids have a shear modulus; gases do not, so they carry only longitudinal (compression) waves.

(e) A pulse is a mixture of many wavelengths. In a dispersive medium the speed depends on wavelength, so the components travel at different speeds and drift apart, changing the shape of the pulse.

(a) Max compression at displacement nodes (b) echo-location (c) different overtones (timbre) (d) gases have no shear modulus (e) different wavelengths travel at different speeds.