Define simple harmonic motion .

A particle moves in simple harmonic motion when its acceleration is always directed towards a fixed point on its line of motion, and is proportional to its displacement from that point. The result is an oscillation back and forth about that fixed point, which is called the centre of oscillation .

True or False? A particle in simple harmonic motion keeps oscillating with the same amplitude for ever.

True. Nothing in the model removes energy from the particle, so the amplitude is fixed once and for all by the conditions at the start. That is a property of the model rather than of the world: real oscillations do lose energy, which is what damping is added to describe.

A LevelFurther MathsEdexcelCore PureFlashcardsDifferential EquationsSimple Harmonic Motion

Simple Harmonic Motion (Edexcel A Level Further Maths: Core Pure): Flashcards

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FrontSimple Harmonic Motion
Define simple harmonic motion.

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Define simple harmonic motion.
A particle moves in simple harmonic motion when its acceleration is always directed towards a fixed point on its line of motion, and is proportional to its displacement from that point.
The result is an oscillation back and forth about that fixed point, which is called the centre of oscillation.
Complete the standard equation of simple harmonic motion:
$\frac{d^{2} x}{d t^{2}} =______$
The completed equation is:
$\frac{d^{2} x}{d t^{2}} = - ω^{2} x$
The minus sign is what turns the acceleration back towards the centre of oscillation, and $x$ is the displacement measured from that point.
Why is the constant in the simple harmonic motion equation written as $ω^{2}$ ?
Writing it as a square guarantees the constant is positive, which is what makes the motion an oscillation rather than a runaway.
It also keeps the solution tidy, since $ω$ itself rather than a square root then appears in the answer.
What do the dots mean in $\ddot{x}$ and $\dot{x}$ ?
Each dot stands for one differentiation with respect to time, so $\dot{x} = \frac{d x}{d t}$ is the velocity and $\ddot{x} = \frac{d^{2} x}{d t^{2}}$ is the acceleration.
Newton's dot notation is used throughout simple harmonic motion, so the defining equation is often written $\ddot{x} = - ω^{2} x$ .
What is the general solution of $\ddot{x} = - ω^{2} x$ , and where does it come from?
The general solution is $x = A \cos ω t + B \sin ω t$ .
Rearranged as $\ddot{x} + ω^{2} x = 0$ the auxiliary equation is $m^{2} + ω^{2} = 0$ , whose roots $m = \pm ω i$ are purely imaginary, so the solution carries no exponential factor.
True or False?
A particle in simple harmonic motion keeps oscillating with the same amplitude for ever.
True.
Nothing in the model removes energy from the particle, so the amplitude is fixed once and for all by the conditions at the start.
That is a property of the model rather than of the world: real oscillations do lose energy, which is what damping is added to describe.
Why is $x = A \cos ω t + B \sin ω t$ often rewritten as $x = R \sin (ω t + α)$ ?
The single-term form makes the amplitude and the phase visible at once, which the two-term form hides.
Here $R = \sqrt{A^{2} + B^{2}}$ is the amplitude and $α$ is the phase constant, found from $\tan α = \frac{A}{B}$ .
What is the period of $x = R \sin (ω t + α)$ , and what does it measure?
The period is $\frac{2 π}{ω}$ , which is the time the particle takes to complete one full oscillation.
Notice that it depends only on $ω$ , so changing the amplitude does not change how long an oscillation takes.
What does the phase constant $α$ tell you about where the particle starts?
The phase constant fixes the position at $t = 0$ , since $x = R \sin α$ when $t = 0$ .
So $α = 0$ starts the particle at the centre of oscillation, while $α = \frac{π}{2}$ starts it at its maximum displacement.
Complete the relation connecting velocity and displacement, where $R$ is the amplitude:
$v^{2} = ω^{2} (______- x^{2})$
The completed relation is:
$v^{2} = ω^{2} (R^{2} - x^{2})$
This form contains no $t$ at all, so it links speed directly to position and is the one to reach for whenever time is not mentioned.
How is the velocity-displacement relation obtained from $\ddot{x} = - ω^{2} x$ ?
Write the acceleration as $v \frac{d v}{d x}$ using the chain rule, which turns the equation into $v \frac{d v}{d x} = - ω^{2} x$ .
Separating the variables and integrating gives $\frac{1}{2} v^{2} = - \frac{1}{2} ω^{2} x^{2} + c$ , and knowing that $v = 0$ at maximum displacement fixes $c$ .
How does the damped harmonic motion equation differ from the simple harmonic motion equation?
A term in the velocity is added, giving $\frac{d^{2} x}{d t^{2}} + k \frac{d x}{d t} + ω^{2} x = 0$ with $k > 0$ .
That term stands for a resistance such as friction or air resistance, and $k$ measures how strong it is.
Complete the conditions separating the three kinds of damping:
heavy damping when $k^{2}______4 ω^{2}$ , critical damping when $k^{2} = 4 ω^{2}$ , and light damping when $k^{2}______4 ω^{2}$ instead.
The completed conditions are:
heavy damping when $k^{2} > 4 ω^{2}$ , critical damping when $k^{2} = 4 ω^{2}$ , and light damping when $k^{2} < 4 ω^{2}$ instead.
It is $k^{2}$ that is compared with $4 ω^{2}$ , which is the discriminant of the auxiliary equation $m^{2} + k m + ω^{2} = 0$ in disguise.
True or False?
Heavier damping always returns a particle to the centre more quickly.
False.
Critical damping is the quickest: for a given $ω^{2}$ , no heavily damped case returns to zero faster than the critical one does.
Past the critical value the extra resistance holds the particle back rather than hurrying it along.
Complete the statement about which kind of damping still oscillates:
Only $______$ damping produces an oscillation, because its auxiliary equation has $______$ roots and so keeps its cosine and sine terms.
The completed statement is:
Only light damping produces an oscillation, because its auxiliary equation has complex roots and so keeps its cosine and sine terms.
The particle then oscillates with an amplitude shrinking towards zero, while heavy and critical damping die away without oscillating at all.
In critical damping the particle returns to the centre without oscillating. Can it ever change direction on the way?
Yes, it can change direction once, depending on the values of $A$ and $B$ in $x = (A + B t) e^{α t}$ .
One change of direction is not an oscillation, since an oscillation means crossing the centre repeatedly rather than turning back a single time.
Why does every damped solution decay to zero, whichever kind of damping it is?
Every root of $m^{2} + k m + ω^{2} = 0$ has a negative real part when $k$ and $ω^{2}$ are both positive, so every exponential in the solution is a decaying one.
The three cases differ only in how the particle gets to zero, not in whether it gets there.
How does the forced harmonic motion equation differ from the damped one?
A driving term is added on the right, giving $\frac{d^{2} x}{d t^{2}} + k \frac{d x}{d t} + ω^{2} x = f (t)$ .
That makes it non-homogeneous, so a particular integral has to be found as well as the complementary function.
In forced motion with damping present, what becomes of the complementary function and the particular integral as $t$ grows large?
The complementary function decays to zero and only the particular integral survives, so the long-term motion is set by the driving force and takes the driving period rather than the natural one.
This depends on the damping being there: with $k = 0$ nothing decays, and the natural oscillation stays in the motion for ever.

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