A periodic motion repeats itself after equal intervals of time. The oscillation of a freely suspended magnet repeats, and a rotating hydrogen molecule returns to the same orientation after each revolution. A single return trip by a swimmer is only one event, not a repeated motion. An arrow released from a bow follows a non-repeating projectile path.
(b) and (c) represent periodic motion.
A single sine or cosine with constant angular frequency represents SHM. Thus $\sin\omega t-\cos\omega t=\sqrt2\sin(\omega t-\pi/4)$ is SHM with period $2\pi/\omega$, and $3\cos(\pi/4-2\omega t)$ is SHM with period $2\pi/(2\omega)=\pi/\omega$. The function $\sin^3\omega t=(3\sin\omega t-\sin3\omega t)/4$ is periodic but contains more than one harmonic component, so it is not a single SHM. Similarly $\cos\omega t+\cos3\omega t+\cos5\omega t$ is periodic with common period $2\pi/\omega$ but is not one SHM. The exponential and polynomial expressions do not repeat.
(a) SHM, period $2\pi/\omega$; (b) periodic but not SHM, period $2\pi/\omega$; (c) SHM, period $\pi/\omega$; (d) periodic but not SHM, period $2\pi/\omega$; (e) non-periodic; (f) non-periodic.
The mean position is the midpoint of AB. In SHM, $a=-\omega^2x$ and $F=ma$, where $x$ is measured from the mean position. At A, $x$ is negative, so $a$ and $F$ are positive; at B, $x$ is positive, so $a$ and $F$ are negative. At the midpoint, $x=0$, so $a=F=0$, and the velocity sign is set by the direction of motion. Points closer to B have positive $x$ and hence negative acceleration and force; points closer to A have negative $x$ and hence positive acceleration and force.
(a) $v=0$, $a$ positive, $F$ positive; (b) $v=0$, $a$ negative, $F$ negative; (c) $v$ negative, $a=0$, $F=0$; (d) $v$ negative, $a$ negative, $F$ negative; (e) $v$ positive, $a$ positive, $F$ positive; (f) $v$ negative, $a$ negative, $F$ negative.
For SHM, acceleration must be proportional to displacement and opposite in direction: $a=-\omega^2x$. Option (c) has this form with $\omega^2=10$. Option (a) has the wrong sign, while (b) and (d) are nonlinear in $x$.
Only (c), $a=-10x$, involves simple harmonic motion.
At $t=0$, $A\cos\phi=1$. Also $v=-A\omega\sin\phi$, and the given initial velocity is $+\omega\,\text{cm s}^{-1}$, so $-A\sin\phi=1$. Squaring and adding gives $A^2=2$, hence $A=\sqrt2\,\text{cm}$. Then $\cos\phi=1/\sqrt2$ and $\sin\phi=-1/\sqrt2$, so $\phi=-\pi/4$. For $x=B\sin(\omega t+\alpha)$, $B\sin\alpha=1$ and $B\cos\alpha=1$, giving $B=\sqrt2\,\text{cm}$ and $\alpha=\pi/4$.
For $x=A\cos(\omega t+\phi)$, $A=\sqrt2\,\text{cm}$ and $\phi=-\pi/4$ (equivalently $7\pi/4$). For $x=B\sin(\omega t+\alpha)$, $B=\sqrt2\,\text{cm}$ and $\alpha=\pi/4$.
A load of $50\,\text{kg}$ stretches the spring by $0.20\,\text{m}$, so $k=50g/0.20=2450\,\text{N m}^{-1}$. For a mass-spring oscillator, $T=2\pi\sqrt{m/k}$, so $m=kT^2/(4\pi^2)=2450(0.6)^2/(4\pi^2)=22.3\,\text{kg}$. Hence weight $mg=22.3\times9.8=2.19\times10^2\,\text{N}$.
The weight is about $2.19\times10^2\,\text{N}$, corresponding to a mass of about $22.3\,\text{kg}$.
Here $k=1200\,\text{N m}^{-1}$, $m=3\,\text{kg}$ and $A=2.0\,\text{cm}=0.020\,\text{m}$. The angular frequency is $\omega=\sqrt{k/m}=\sqrt{1200/3}=20\,\text{rad s}^{-1}$. Thus $f=\omega/(2\pi)=20/(2\pi)=3.18\,\text{Hz}$. Maximum acceleration is $\omega^2A=20^2(0.020)=8.0\,\text{m s}^{-2}$, and maximum speed is $\omega A=20(0.020)=0.40\,\text{m s}^{-1}$.
(i) $3.18\,\text{Hz}$; (ii) $8.0\,\text{m s}^{-2}$; (iii) $0.40\,\text{m s}^{-1}$.
From Exercise 13.9, $A=0.020\,\text{m}$ and $\omega=20\,\text{rad s}^{-1}$. If the body is at the mean position at $t=0$ and moves initially in the positive direction, $x=A\sin\omega t=0.020\sin20t$. At maximum stretch, $x=A$ at $t=0$, so $x=A\cos\omega t=0.020\cos20t$. At maximum compression, $x=-A$ at $t=0$, so $x=-A\cos\omega t=-0.020\cos20t$. These are phase-shifted forms of the same SHM.
Taking motion through the mean position initially towards positive $x$: (a) $x=0.020\sin20t\,\text{m}$; (b) $x=0.020\cos20t\,\text{m}$; (c) $x=-0.020\cos20t\,\text{m}$. The frequency and amplitude are the same; only the initial phase differs.
The spring tension is $F$ in each case, so Hooke's law gives extension $x=F/k$ for both arrangements. In Fig. 13.21(a), the fixed-end spring with mass $m$ has period $T=2\pi\sqrt{m/k}$. In Fig. 13.21(b), the centre of mass remains fixed and the two equal masses move symmetrically. For the relative extension coordinate, the reduced mass is $m/2$, so $T=2\pi\sqrt{(m/2)/k}=2\pi\sqrt{m/(2k)}$.
(a) The maximum extension is $F/k$ in both cases. (b) The periods are $2\pi\sqrt{m/k}$ for Fig. 13.21(a) and $2\pi\sqrt{m/(2k)}$ for Fig. 13.21(b).
The stroke is $2A=1.0\,\text{m}$, so $A=0.50\,\text{m}$. The angular frequency is $200\,\text{rad min}^{-1}=200/60=3.33\,\text{rad s}^{-1}$. Maximum speed in SHM is $v_{\max}=\omega A=3.33\times0.50=1.67\,\text{m s}^{-1}$.
$1.67\,\text{m s}^{-1}$.
For the same pendulum, $T=2\pi\sqrt{l/g}$, so $T_m/T_e=\sqrt{g_e/g_m}$. Therefore $T_m=3.5\sqrt{9.8/1.7}=8.4\,\text{s}$.
$8.4\,\text{s}$ approximately.
In the non-inertial frame of the car, the bob experiences gravity $Mg$ downward and a centrifugal force $Mv^2/R$ radially outward. The effective acceleration is the vector sum, $g_{\text{eff}}=\sqrt{g^2+(v^2/R)^2}$. For small oscillations about the equilibrium direction, the pendulum period is $T=2\pi\sqrt{l/g_{\text{eff}}}=2\pi\sqrt{l/\sqrt{g^2+v^4/R^2}}$.
$T=2\pi\sqrt{\dfrac{l}{\sqrt{g^2+v^4/R^2}}}$.
Suppose one limb rises by $y$ and the other falls by $y$ from the equilibrium level. The level difference is then $2y$, producing pressure difference $\rho g(2y)$. The restoring force on cross-sectional area $A$ is $F=-2\rho gAy$. If $L$ is the total length of mercury in the tube, the moving mass is $\rho AL$. Thus $\rho AL\ddot{y}=-2\rho gAy$, or $\ddot{y}+(2g/L)y=0$. This is the equation of SHM, with angular frequency $\sqrt{2g/L}$ and period $2\pi\sqrt{L/(2g)}$.
After the pump is removed, the mercury column executes SHM with period $T=2\pi\sqrt{L/(2g)}$, where $L$ is the total length of the mercury column.