Periodic Motion

(d) $f/\sqrt{2}$

When the block moves by $x$, one spring is stretched and the other compressed, and both push it back towards the mean position. So the restoring force is $F = -2kx$ and

$$f = \frac{1}{2\pi}\sqrt{\frac{2k}{m}}$$

With one spring removed, $F = -kx$, so $f' = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} = \dfrac{f}{\sqrt2}$.

(a) Basic equation of SHM

In SHM the acceleration is directly proportional to the displacement from the mean position and is always directed towards it:

$$a = \frac{d^2x}{dt^2} = -\omega^2 x \qquad \text{or} \qquad \frac{d^2x}{dt^2} + \omega^2 x = 0$$

Its solution is $x = A\sin(\omega t + \phi)$, where $A$ is the amplitude, $\omega$ the angular frequency and $\phi$ the initial phase.

(b) Velocity and acceleration in SHM

Take the displacement as

$$x = A\sin\omega t$$

Velocity:

$$v = \frac{dx}{dt} = A\omega\cos\omega t = A\omega\sqrt{1 - \sin^2\omega t} = A\omega\sqrt{1 - \frac{x^2}{A^2}}$$ $$\boxed{v = \pm\,\omega\sqrt{A^2 - x^2}}$$

The speed is maximum, $v_{\max} = A\omega$, at the mean position ($x = 0$) and zero at the extremes ($x = \pm A$).

Acceleration:

$$a = \frac{dv}{dt} = -A\omega^2\sin\omega t$$ $$\boxed{a = -\omega^2 x}$$

The acceleration is zero at the mean position and maximum, $a_{\max} = \omega^2 A$, at the extremes.

(c) Tuning fork

Given: frequency $f = 550$ vibrations per second $= 550$ Hz.

Angular frequency:

$$\omega = 2\pi f = 2\times 3.1416 \times 550 \approx \boxed{3.46\times10^{3}\ \text{rad s}^{-1}}$$

Time period:

$$T = \frac{1}{f} = \frac{1}{550} \approx \boxed{1.82\times10^{-3}\ \text{s}}$$

(b) $T_2 = 2T_1$

Take $x = A\sin\omega t$, with $\omega = 2\pi/T$. For $x = A/2$: $\sin\omega t = \tfrac12$, so $\omega t = \pi/6$ and $T_1 = T/12$. Reaching $x = A$ takes $T/4$, so $T_2 = \dfrac{T}{4} - \dfrac{T}{12} = \dfrac{T}{6} = 2T_1$. The particle slows down as it nears the extreme, so the second half takes longer.

(a) Simple harmonic motion

SHM is a periodic to-and-fro motion in which the restoring force (and so the acceleration) is directly proportional to the displacement from the mean position and is always directed towards the mean position: $F = -kx$, or $a = -\omega^2 x$.

(b) Simple pendulum performs SHM

   ==========S==========   rigid support
             |\
             | \
             |θ \  l (string, tension T)
             |   \
             |    \
             |     (m)  bob
             |      |
             |      v  mg
             O
       mean position

   x = displacement of the bob along the arc from O
   mg cos θ : along the string, balanced by T
   mg sin θ : along the arc towards O (restoring force)

1. A bob of mass $m$ hangs from a light, inextensible string of length $l$. When displaced through a small angle $\theta$, its weight $mg$ has two components:
   • $mg\cos\theta$ along the string, balanced by the tension $T$;
   • $mg\sin\theta$ along the arc, towards the mean position. This is the restoring force.
2. So

$$F = -mg\sin\theta$$

(the minus sign shows it acts opposite to the displacement). 3. For small $\theta$ (in radians), $\sin\theta \approx \theta = \dfrac{x}{l}$, where $x$ is the displacement of the bob along the arc. Then

$$F = -\frac{mg}{l}x$$

4. Acceleration:

$$a = \frac{F}{m} = -\frac{g}{l}x$$

5. Since $g/l$ is constant, $a \propto -x$: the acceleration is proportional to the displacement and directed towards the mean position. Hence the motion is simple harmonic.
6. Comparing with $a = -\omega^2 x$:

$$\omega^2 = \frac{g}{l} \;\Rightarrow\; \omega = \sqrt{\frac{g}{l}}$$

7. Time period:

$$T = \frac{2\pi}{\omega} \;\Rightarrow\; \boxed{T = 2\pi\sqrt{\frac{l}{g}}}$$

(c) Factors affecting the time period

From $T = 2\pi\sqrt{l/g}$, the time period depends on:

• the effective length $l$ ($T \propto \sqrt l$), and
• the acceleration due to gravity $g$ at that place ($T \propto 1/\sqrt g$).

It does not depend on the mass of the bob or (for small swings) on the amplitude.

(d) $\dfrac{T}{12}$

Starting from the mean position, $x = A\sin\omega t$. For $x = A/2$: $\sin\omega t = \tfrac12$, so $\omega t = \dfrac{\pi}{6}$ and

$$t = \frac{\pi}{6}\cdot\frac{T}{2\pi} = \frac{T}{12}$$

a) Acceleration of a pendulum bob is proportional to displacement

1. Let a bob of mass $m$ hang from a string of length $l$ and be displaced through a small angle $\theta$, so that its displacement along the arc is $x$.
2. The weight $mg$ has a component $mg\cos\theta$ along the string (balanced by tension) and a component $mg\sin\theta$ along the arc, towards the mean position. This is the restoring force:

$$F = -mg\sin\theta$$

3. For small $\theta$, $\sin\theta \approx \theta = x/l$, so

$$F = -\frac{mg}{l}x \;\Rightarrow\; a = \frac{F}{m} = -\frac{g}{l}x$$

4. As $g$ and $l$ are constant,

$$\boxed{a \propto -x}$$

So the acceleration is directly proportional to the displacement and directed towards the mean position (the motion is SHM).

b) Numerical

Given:

• Mass, $m = 8$ kg
• Amplitude, $A = 30$ cm $= 0.30$ m
• Restoring force $= 60$ N. Taken as the maximum restoring force, which acts at the extreme position ($x = A$).
• Displacement for part (iii), $x = 12$ cm $= 0.12$ m

Force constant:

$$F_{\max} = kA \;\Rightarrow\; k = \frac{60}{0.30} = 200\ \text{N m}^{-1}$$

i) Time period

$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{8}{200}} = 2\pi\times0.2$$ $$\boxed{T \approx 1.26\ \text{s}}$$

ii) Maximum acceleration

$$a_{\max} = \frac{F_{\max}}{m} = \frac{60}{8} = \boxed{7.5\ \text{m s}^{-2}}$$

iii) Kinetic energy at $x = 0.12$ m

$$K = \tfrac12 k(A^2 - x^2) = \tfrac12\times200\times\left[(0.30)^2 - (0.12)^2\right] = 100\times(0.09 - 0.0144)$$ $$\boxed{K \approx 7.56\ \text{J}}$$

(D) 31.4 cm s$^{-1}$

From the graph, amplitude $A = 1$ cm and one full cycle takes $T = 0.2$ s. So $\omega = \dfrac{2\pi}{T} = 10\pi$ rad s$^{-1}$, and

$$v_{\max} = A\omega = 1 \times 10\pi \approx 31.4\ \text{cm s}^{-1}$$

a) i) Simple harmonic motion

SHM is a periodic to-and-fro motion in which the acceleration of the body is directly proportional to its displacement from the mean position and is always directed towards that position: $a = -\omega^2 x$.

a) ii) Maximum acceleration and time period

Given:

• Spring constant, $k = 18$ N m$^{-1}$
• Mass, $m = 0.15$ kg
• Extra extension on pulling, which is the amplitude, $A = 4$ cm $= 0.04$ m

Angular frequency:

$$\omega^2 = \frac{k}{m} = \frac{18}{0.15} = 120\ \text{s}^{-2}, \qquad \omega = \sqrt{120} \approx 10.95\ \text{rad s}^{-1}$$

Maximum acceleration (at the extreme position):

$$a_{\max} = \omega^2 A = 120\times0.04 = \boxed{4.8\ \text{m s}^{-2}}$$

Time period:

$$T = 2\pi\sqrt{\frac{m}{k}} = \frac{2\pi}{10.95} \approx \boxed{0.574\ \text{s}}$$

b) Total energy and energy graph

At displacement $x$:

$$\text{P.E.} = \tfrac12 kx^2 = \tfrac12 m\omega^2x^2, \qquad \text{K.E.} = \tfrac12 m\omega^2(A^2 - x^2)$$ $$\boxed{E = \text{K.E.} + \text{P.E.} = \tfrac12 m\omega^2A^2 = \tfrac12 kA^2}$$

This is constant: it does not depend on $x$.

 Energy        ---- total energy E (constant)
   ^           k    kinetic energy
   |           p    potential energy
 E |p---------------kkkkkkkkk---------------p
   | p          kkkk         kkkk          p
   |  pp      kk                 kk      pp
   |    p   kk                     kk   p
   |     pkk                         kkp
   |     kpp                         ppk
   |    k   pp                     pp   k
   |  kk      pp                 pp      kk
   | k          pppp         pppp          k
   |k               ppppppppp               k
   +-----------------------------------------> x
  -A                   0                    +A

• P.E. ($\tfrac12 kx^2$) is a parabola: zero at $x = 0$, maximum ($= E$) at $x = \pm A$.
• K.E. ($\tfrac12 k(A^2 - x^2)$) is an inverted parabola: maximum ($= E$) at $x = 0$, zero at $x = \pm A$.
• At every point, K.E. + P.E. $= E$ (the horizontal line).

(A) $0.2\pi$ m/s

From the graph, amplitude $A = 0.20$ m and period $T = 2.0$ s, so $\omega = \dfrac{2\pi}{T} = \pi$ rad s$^{-1}$. The speed is greatest at the mean position:

$$v_{\max} = A\omega = 0.20\times\pi = 0.2\pi\ \text{m s}^{-1}$$

a) Total energy in SHM is constant

Consider a particle of mass $m$ in SHM with amplitude $A$ and angular frequency $\omega$, at displacement $x$.

1. Kinetic energy. The velocity is $v = \omega\sqrt{A^2 - x^2}$, so

$$\text{K.E.} = \tfrac12 mv^2 = \tfrac12 m\omega^2(A^2 - x^2)$$

2. Potential energy. The restoring force is $F = -m\omega^2 x$. The work done against it to move the particle from $0$ to $x$ is stored as P.E.:

$$\text{P.E.} = \int_0^x m\omega^2x\,dx = \tfrac12 m\omega^2x^2$$

3. Total energy.

$$E = \tfrac12 m\omega^2(A^2 - x^2) + \tfrac12 m\omega^2x^2$$ $$\boxed{E = \tfrac12 m\omega^2A^2}$$

4. Here $m$, $\omega$ and $A$ are all constant, so $E$ does not depend on $x$ or on time. The total energy of a particle in SHM is constant. Energy only changes form: all kinetic at the mean position, all potential at the extremes.

b) Graph of P.E. against displacement

P.E. $= \tfrac12 m\omega^2x^2$, so the graph is a parabola with its lowest point (zero) at the mean position and maximum value $E$ at $x = \pm A$.

  P.E.
   ^
 E |*                         *
   | *                       *
   |  *                     *
   |   *                   *
   |     *               *
   |        *         *
   |            * *
   +-------------+-------------->  x
  -A             0            +A

c) Time period on the Moon

$T = 2\pi\sqrt{l/g}$, so $T \propto \dfrac{1}{\sqrt g}$. On the Moon, $g_m = \dfrac{g}{6}$:

$$T_m = T\sqrt{\frac{g}{g_m}} = \sqrt6\,T \approx \boxed{2.45\,T}$$

The pendulum swings more slowly on the Moon.

(b) 10%

$T \propto \sqrt l$. The new length is $1.21\,l$, so $T' = \sqrt{1.21}\,T = 1.1\,T$. The period increases by 10%.

a) Definition and equations

Simple harmonic motion is a periodic to-and-fro motion in which the acceleration (or restoring force) is directly proportional to the displacement from the mean position and is always directed towards the mean position: $a = -\omega^2 x$.

General equation for displacement:

$$x = A\sin(\omega t + \phi)$$

where $A$ is the amplitude, $\omega$ the angular frequency and $\phi$ the initial phase.

Velocity: $v = \dfrac{dx}{dt} = A\omega\cos(\omega t + \phi)$.

General equation for acceleration:

$$a = \frac{dv}{dt} = -A\omega^2\sin(\omega t + \phi) \;\Rightarrow\; \boxed{a = -\omega^2x}$$

b) Graph of velocity in SHM

If $x = A\sin\omega t$, then $v = A\omega\cos\omega t$, a cosine curve of amplitude $A\omega$:

  v
+Aw |****                                 ****
    |    **                             **
    |      **                         **
    |        **                     **
 0  +----------*-------------------*----------> t
    |           **               **
    |             **           **
    |               **       **
-Aw |                 *******
    0         T/4       T/2        3T/4      T
                  (w = angular frequency)

The velocity is maximum ($\pm A\omega$) at the mean position and zero at the extremes. (Plotted against displacement, $v^2/(A\omega)^2 + x^2/A^2 = 1$, which is an ellipse.)

c) Simple pendulum at the centre of the Earth

$T = 2\pi\sqrt{l/g}$. At the centre of the Earth, $g = 0$, so

$$T = 2\pi\sqrt{\frac{l}{0}} \to \infty$$

The time period becomes infinite: there is no restoring force (no weight), so a displaced bob does not return and the pendulum does not oscillate.

(d) Graph D

At the mean position the speed is maximum, so for motion starting there $v = \omega r\cos\omega t$ and

$$K = \tfrac12 mv^2 = \tfrac12 m\omega^2r^2\cos^2\omega t$$

The kinetic energy is therefore maximum at $t=0$ (the curve starts above the axis), never negative, falls to zero at the extreme positions and repeats twice in every period, giving a series of arches that start at a maximum.

(a)(i) Amplitude

The amplitude is the maximum displacement of the particle from its mean position.

(a)(ii) Acceleration

$$v = \frac{dy}{dt} = r\omega\cos\omega t$$ $$a = \frac{dv}{dt} = -r\omega^2\sin\omega t = -\omega^2 y$$

The acceleration is proportional to the displacement and directed towards the mean position, with maximum value $r\omega^2$.

(b) Spring cut in half

The force constant of each half is doubled, $k' = 2k$ (a shorter spring is stiffer, since $k \propto 1/\text{length}$).

The frequency with the same mass is

$$f' = \frac{1}{2\pi}\sqrt{\frac{2k}{m}} = \sqrt{2}\,f$$

so the frequency of the half spring is $\sqrt2$ times that of the whole spring.

(a) straight line

Since $T = 2\pi\sqrt{l/g}$, we have $T^2 = \dfrac{4\pi^2}{g}\,l$, so $l = \dfrac{g}{4\pi^2}T^2$. Thus $l \propto T^2$, and the graph is a straight line through the origin.

(a) Simple harmonic motion is the to-and-fro motion of a particle in which the acceleration is directly proportional to its displacement from the mean position and is always directed towards it: $a = -\omega^2 x$.

(b) At the mean position the velocity is maximum, so $v = v_{max}\cos\omega t$. The graph is a cosine curve. It starts from $+v_{max}$ at $t=0$, falls to zero at $t = T/4$, reaches $-v_{max}$ at $T/2$, returns to zero at $3T/4$ and is back at $+v_{max}$ at $t=T$. (The axes are $v$ vertical and $t$ horizontal.)

(c) Use $v^2 = \omega^2(A^2 - x^2)$ for the two positions:

$64 = \omega^2(A^2 - 144), \qquad 144 = \omega^2(A^2 - 64)$

Subtracting the first from the second:

$80 = 80\,\omega^2 \;\Rightarrow\; \omega = 1\ \text{rad/s}$

Then $A^2 = 64 + 144 = 208$, so $A = 14.4$ cm.

$v_{max} = \omega A = 14.4\ \text{cm/s}, \qquad a_{max} = \omega^2 A = 14.4\ \text{cm/s}^2$

(a) For SHM the acceleration is directly proportional to the displacement from the mean position and is directed opposite to it:

$a = -\omega^2 x$

(b) The time period is

$T = \frac1f = \frac{1}{10^{14}} = 10^{-14}\ \text{s}$

The angular frequency is $\omega = 2\pi f = 2\pi\times10^{14}$ rad/s, so

$v_{max} = \omega A = 2\pi\times10^{14}\times2.0\times10^{-12} \approx 1.26\times10^{3}\ \text{m/s}$

Graph: The displacement is $x = A\sin\omega t$ (or cosine), a sinusoidal curve of amplitude $2.0\times10^{-12}$ m. In one cycle (from $0$ to $10^{-14}$ s) it rises from zero to $+A$ at $T/4$, returns to zero at $T/2$, falls to $-A$ at $3T/4$ and returns to zero at $T$.

(c) No. A pendulum's period is $T = 2\pi\sqrt{l/g_{eff}}$. In a freely falling elevator the effective gravity is $g_{eff}=0$, so the restoring force vanishes, the period becomes infinite and the pendulum does not oscillate. The clock stops and does not keep correct time.

(B) 0.044 J

The total mechanical energy of the pendulum is constant. At the extreme point it is all potential energy (0.044 J); at the mean point it is all kinetic energy, so K.E. $= 0.044$ J.

Check: $E = \tfrac12 m\omega^2A^2 = \tfrac12\times0.40\times\pi^2\times(0.15)^2 \approx 0.044$ J, with $\omega = 2\pi/T = \pi$ rad s$^{-1}$.

(i) Simple harmonic motion

SHM is a periodic to-and-fro motion in which the acceleration of the body is directly proportional to its displacement from the mean position and is always directed towards it: $a = -\omega^2x$.

(ii) Time period of a mass on a vertical spring

   /////////        /////////        /////////
       |                |                |
       >                >                >
       <                <                <
       >                >                >
       <                <                <
       |  natural       >                >
          length        <                <
                        |   extension e  >
                      [ m ] equilibrium  <
                                         |   extra x
                                       [ m ] pulled down
                                             (restoring force -kx)

1. A light spring of force constant $k$ hangs from a rigid support. When a mass $m$ is attached, the spring stretches by $e$ and the mass rests in equilibrium. Then

$$mg = ke \qquad \text{(i)}$$

2. Now pull the mass down a further distance $x$. The total extension is $(e + x)$, so the spring pulls upward with force $k(e + x)$, while the weight $mg$ acts downward.
3. Net (restoring) force, taking downward as positive:

$$F = mg - k(e + x) = mg - ke - kx$$

4. Using (i), $mg - ke = 0$, so

$$F = -kx$$

5. Acceleration:

$$a = \frac{F}{m} = -\frac{k}{m}x$$

Since $k/m$ is constant, $a \propto -x$, so the motion is simple harmonic. 6. Comparing with $a = -\omega^2x$:

$$\omega = \sqrt{\frac{k}{m}}$$

7. Time period:

$$T = \frac{2\pi}{\omega} \;\Rightarrow\; \boxed{T = 2\pi\sqrt{\frac{m}{k}}}$$

(iii) Inside a satellite

The time period stays the same, $T = 2\pi\sqrt{m/k}$. It depends only on $m$ and $k$, not on $g$. In the weightless satellite the spring is not stretched at rest ($e = 0$), but when the mass is displaced the spring force $-kx$ still acts, so it oscillates with the same period.

(Compare a simple pendulum: $T = 2\pi\sqrt{l/g}$ becomes infinite in a satellite, so it does not oscillate there.)

(i) The force $F = -Ax$ produces SHM

Let the vibrating part of the rock have mass $m$. By Newton's second law,

$$ma = -Ax \;\Rightarrow\; a = -\frac{A}{m}x$$

$A$ and $m$ are constants, so we can write $\dfrac{A}{m} = \omega^2$:

$$a = -\omega^2x, \qquad \omega = \sqrt{\frac{A}{m}}$$

The acceleration is directly proportional to the displacement and always directed towards the mean position (minus sign). This is the defining condition of SHM, so the rock vibrates with simple harmonic motion, with period $T = 2\pi\sqrt{m/A}$.

(ii) Speed in SHM: $v = \pm\omega\sqrt{A^2 - x^2}$

(In this part $A$ is the amplitude.)

1. The displacement in SHM is

$$x = A\sin\omega t$$

2. Differentiating with respect to time:

$$v = \frac{dx}{dt} = A\omega\cos\omega t$$

3. Using $\cos\omega t = \pm\sqrt{1 - \sin^2\omega t} = \pm\sqrt{1 - \dfrac{x^2}{A^2}}$:

$$v = \pm A\omega\sqrt{1 - \frac{x^2}{A^2}} = \pm\,\omega\sqrt{A^2 - x^2}$$ $$\boxed{v = \pm\,\omega\sqrt{A^2 - x^2}}$$

(iii) Maximum speed of the building

Given: frequency $f = 21$ Hz, amplitude $A = 0.05$ m.

The speed is maximum at $x = 0$:

$$v_{\max} = A\omega = A(2\pi f) = 0.05\times2\times3.14\times21$$ $$\boxed{v_{\max} \approx 6.6\ \text{m s}^{-1}}$$

(iv) Why tall buildings suffer more from low-frequency S-waves

• Every building has its own natural frequency of vibration. A taller building is like a longer, more flexible oscillator, so its natural frequency is low (its period is long).
• S-waves of an earthquake are mostly of low frequency. For tall buildings, this frequency is close to their natural frequency, so resonance occurs.
• At resonance the building absorbs energy from the waves on every cycle and swings with a very large amplitude. Taller buildings also sway with large displacement at the top. This causes cracks and can make the building collapse.
• Short, stiff buildings have a high natural frequency, far from the S-wave frequency, so they do not resonate and suffer less.