Chapter 13: Oscillations (Oscillations and Waves)

(a) A single return trip is not repeated: not periodic. (b) The magnet oscillates about the N-S direction: periodic. (c) Rotation repeats itself: periodic. (d) The arrow does not return: not periodic.

(b) and (c)

(a) Periodic, but not a to-and-fro motion about a mean position: not SHM.

(b) The restoring force is proportional to the displacement of the levels: SHM.

(c) For small displacements the restoring force ∝ displacement: (nearly) SHM.

(d) A polyatomic molecule has several natural frequencies; its general vibration is a superposition of SHMs of different frequencies: periodic but not SHM.

SHM: (b), (c). Periodic but not SHM: (a), (d).

(a) Never repeats: not periodic.

(b) The same pattern repeats every 2 s: periodic, T = 2 s.

(c) Only some positions recur; the whole motion does not repeat: not periodic.

(d) Repeats every 2 s: periodic, T = 2 s.

(b) and (d) are periodic, each with T = 2 s; (a) and (c) are not.

(a) $\sin \omega t-\cos \omega t=\sqrt{2}\sin \left(\omega t-\frac{\pi }{4}\right)$: SHM, $T=\frac{2\pi }{\omega }$.

(b) ${\sin }^3\omega t=\frac{1}{4}(3\sin \omega t-\sin 3\omega t)$: two SHMs of different frequencies: periodic, not SHM, $T=\frac{2\pi }{\omega }$.

(c) $3\cos (2\omega t-\frac{\pi }{4})$: SHM, $T=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$.

(d) Sum of SHMs with ω, 3ω, 5ω: periodic, not SHM, $T=\frac{2\pi }{\omega }$ (the longest period).

(e) Decreases steadily to zero: non-periodic.

(f) Increases without limit: non-periodic (and not physically acceptable as t → ∞).

SHM: (a) T = 2π/ω, (c) T = π/ω. Periodic not SHM: (b), (d), T = 2π/ω. Non-periodic: (e), (f).

Take the mid-point O as origin: A at x = −5 cm, B at x = +5 cm. Acceleration and force are $\propto -x$ (towards O); velocity sign is the direction of motion.

(a) At A: v = 0; x < 0, so a, F > 0. (b) At B: v = 0; a, F < 0. (c) At O towards A: v < 0; x = 0, so a = F = 0. (d) x = +3 towards A: v < 0, a < 0, F < 0. (e) x = −2 towards B: v > 0, a > 0, F > 0. (f) x = +1 towards A: v < 0, a < 0, F < 0.

(v, a, F): (a) 0, +, + (b) 0, −, − (c) −, 0, 0 (d) −, −, − (e) +, +, + (f) −, −, −

SHM needs $a=-{\omega }^{2}x$: proportional to x and opposite to it. Only (c) qualifies (ω = √10 rad s⁻¹).

(c)

Cosine form: $A\cos \varphi =1$ and $v(0)=-A\omega \sin \varphi =\omega \Rightarrow A\sin \varphi =-1$.

$A=\sqrt{1+1}=\sqrt{2}$ cm; $\tan \varphi =-1$ with $\cos \varphi >0$: $\varphi =-\frac{\pi }{4}$ (or $\frac{7\pi }{4}$).

Sine form: $B\sin \alpha =1$ and $B\omega \cos \alpha =\omega \Rightarrow B\cos \alpha =1$. So $B=\sqrt{2}$ cm, $\alpha =\frac{\pi }{4}$.

A = √2 cm, φ = −π/4 (7π/4); B = √2 cm, α = π/4

$k=\frac{50\times 9.8}{0.20}=2450{\text{N m}}^{-1}$. $T=2\pi \sqrt{\frac{m}{k}}\Rightarrow m=\frac{k{T}^2}{4{\pi }^2}=\frac{2450\times 0.36}{39.5}=22.3$ kg.

Weight $=22.3\times 9.8=219$ N.

About 219 N

$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{1200}{3}}=20{\text{rad s}}^{-1}$, $A=0.020$ m.

(i) $\nu =\frac{\omega }{2\pi }=3.2$ Hz (ii) $a_{\max }={\omega }^{2}A=400\times 0.020=8.0{\text{m s}}^{-2}$ (iii) $v_{\max }=\omega A=0.40{\text{m s}}^{-1}$

(i) 3.2 s⁻¹ (ii) 8.0 m s⁻² (iii) 0.4 m s⁻¹

ω = 20 rad s⁻¹, A = 2 cm.

(a) x = 0 at t = 0 (moving right): $x=2\sin 20t$. (b) x = +2 at t = 0: $x=2\cos 20t$. (c) x = −2 at t = 0: $x=-2\cos 20t$.

They have the same amplitude and frequency; they differ only in initial phase.

(a) x = 2 sin 20t (b) x = 2 cos 20t (c) x = −2 cos 20t (cm); differ only in initial phase

(a) ω = 2π/2 = π rad s⁻¹. P starts at angle −90° and moves clockwise, so its angle is $-\frac{\pi }{2}-\pi t$:

$x=3\cos \left(-\pi t-\frac{\pi }{2}\right)=3\cos \left(\pi t+\frac{\pi }{2}\right)=-3\sin \pi t$ (cm).

(b) ω = 2π/4 = π/2 rad s⁻¹. P starts at angle 180°: $x=2\cos \left(\frac{\pi t}{2}+\pi \right)=-2\cos \frac{\pi t}{2}$ (the sense of rotation does not change the x-projection here).

(a) x = −3 sin πt cm (b) x = −2 cos (πt/2), in metres as in the figure

Write each as $x=A\cos (\omega t+\varphi )$; then the circle has radius A, angular speed ω, and P starts at angle φ from the +x axis (anticlockwise).

(a) $-2\sin (3t+\frac{\pi }{3})=2\cos (3t+\frac{\pi }{3}+\frac{\pi }{2})=2\cos (3t+\frac{5\pi }{6})$: radius 2 cm, ω = 3 rad s⁻¹, start at 150°.

(b) $\cos (\frac{\pi }{6}-t)=\cos (t-\frac{\pi }{6})$: radius 1 cm, ω = 1 rad s⁻¹, start at −30°.

(c) $3\sin (2\pi t+\frac{\pi }{4})=3\cos (2\pi t-\frac{\pi }{4})$: radius 3 cm, ω = 2π rad s⁻¹, start at −45°.

(d) $2\cos \pi t$: radius 2 cm, ω = π rad s⁻¹, start at 0° (on the +x axis).

(a) r = 2 cm, ω = 3, φ = 5π/6 (b) r = 1 cm, ω = 1, φ = −π/6 (c) r = 3 cm, ω = 2π, φ = −π/4 (d) r = 2 cm, ω = π, φ = 0

(a) In both cases the tension in the spring is F, so the extension is $\frac{F}{k}$.

(b) (a): $T=2\pi \sqrt{\frac{m}{k}}$. (b): both masses oscillate symmetrically about the fixed centre of the spring; each half of the spring has constant 2k, so $T=2\pi \sqrt{\frac{m}{2k}}$ (equivalently, reduced mass $\frac{m}{2}$ on spring k).

(a) F/k in both (b) 2π√(m/k) and 2π√(m/2k)

$A=0.5$ m; $v_{\max }=A\omega =0.5\times 200$.

100 m/min

$T\propto \frac{1}{\sqrt{g}}$: $T_m=3.5\sqrt{\frac{9.8}{1.7}}=3.5\times 2.40$.

About 8.4 s

In the car's frame the bob feels gravity g downwards and a centrifugal acceleration $\frac{v^2}{R}$ horizontally. The effective acceleration is $g_{\text{eff}}=\sqrt{g^2+\frac{v^4}{R^2}}$.

$T=2\pi \sqrt{\frac{l}{\sqrt{g^2+v^4/R^2}}}$

In equilibrium, weight = upthrust. If the cork is pushed down a further distance x, the extra upthrust is $Ax{\rho }_{1}g$ upwards, so the restoring force $F=-(A{\rho }_{1}g)x$: SHM with $k=A{\rho }_{1}g$.

Mass $m=Ah\rho$. $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{Ah\rho }{A{\rho }_{1}g}}=2\pi \sqrt{\frac{h\rho }{{\rho }_{1}g}}$.

Restoring force ∝ −x, so SHM with T = 2π√(hρ/ρ₁g). Shown.

When the pump is removed, both ends are at atmospheric pressure. If the level in one arm is y above the equilibrium and the other y below, the difference in levels is 2y. The unbalanced weight $2yA\rho g$ acts to restore the levels: $F=-(2A\rho g)y$.

The restoring force is proportional to the displacement and opposite to it, so the motion is simple harmonic. With total column length L (mass $AL\rho$): $T=2\pi \sqrt{\frac{L}{2g}}$.

F = −2Aρg·y ∝ −y: SHM (T = 2π√(L/2g)).