ch11 Flashcards
conditions for simple harmonic motion
force is directly proportional to displacement, and the acceleration (force) acts in the opposite direction to the displacement this tells us force is always directed to the position of equilibrium
examples of simple harmonic oscillators
springs and pendulums
what is the time for one complete swing of an oscillator
time period
what is the amplitude of an oscillation
greatest displacement from the equilibrium position
what is simple harmonic motion
an oscillation in which the restoring force on an object (acceleration of the object) is directly proportional to it’s displacement from the centre, and is directed towards the centre
equation for angular frequency, w, rate of rotation
w = 2πf rad s^-1
versions of equation for displacement applying angular velocity disp = max at t = 0 disp = 0 at t = 0
if displacement is at max when t = 0 use the equation x = Acos(2πft) -> x = Acos(wt) if displacement is 0 when t = 0 x = Asin(2πft) -> x = Asin(wt) x = displacement in metres A = amplitude t = time in seconds
how to calculate acceleration using angular velocity and the differentiation model for disp to accel
a = -w²x = -4π²f²x = d²x / dt²
what does a negative value mean when you calculate displacement of an object using the displacement equation x = Acos(2πft)
the mass is below the equilibrium at that point in instant
what is the acceleration at the point in equilibrium and why is this
acceleration at the point of equilibrium is 0, as x = 0 and a ∝ -x
a simple harmonic oscillator has an amplitude of 0.02m, the frequency of the oscillation is 1.5Hz calculate the maximum acceleration
maximum acceleration is at max displacement as a ∝ -x so x is 0.02 a = -4 * π² * f² * 0.02 = -1.8 ms^-2
displacement, x, against time graph for SH oscillator max disp at t = 0 T = 2π what type of graph is it
when amplitude A is max (max displacement) at t = 0, it’s a cosine graph
velocity, v, against time graph for SH oscillator max disp at t = 0 T = 2π what type of graph is it
the graph of velocity is the gradient of the displacement graph at t = 0, disp graph has max displacement so velocity starts at 0 the graph of velocity is a reflection of sin x in the x-axis (-sinx)
acceleration, a, against time graph for Sh oscillator max disp at t = 0 T = 2π what type of graph is it
the graph of acceleration is the gradient of the velocity graph at t = 0, displacement graph has max displacement so velocity is 0 and as velocity is 0, acceleration starts at minimum value the graph of acceleration is a reflection of cos x in the x-axis (-cosx) the graph of acceleration is also the opposite of the displacement graph
displacement, x, against time graph for SH oscillator min displacement at t = 0 T = 2π what type of graph is it
at t = 0, displacement is 0 (0 amplitude A) it’s a normal sin x graph
velocity, v, against time graph for SH oscillator min displacement at t = 0 T = 2π what type of graph is it
the graph of velocity is the gradient of the displacement graph at t = 0, disp graph has 0 displacement so velocity starts at max value it’s a cosine graph
acceleration, a, against time graph for SH oscillator min displacement at t = 0 T = 2π what type of graph is it
the graph of acceleration is the gradient of the velocity graph at t = 0, displacement graph has 0 displacement so velocity is at max acceleration graph is the opposite of the displacement graph, but as displacement is 0, acceleration starts at 0 a = -x acceleration is a reflection of the displacement graph
value of velocity when displacement is maximum
0
value of velocity when displacement is 0
maximum
value of displacement when velocity is 0
maximum
value of displacement when velocity is maximum
0
what equation can be used to calculate acceleration, and thus velocity and displacement, from iterative models and calculations
F = ma
maximum value of acceleration graph
w²A
maximum value of velocity graph
wA
maximum value of displacement graph
A
equation for oscillation of a mass on a spring with a restoring force (acceleration to the equilibrium) proportional to displacement
a = -kx / m = Δ²x / Δt²
equation for force exerted when a mass is pushed between springs
F = kx
F = force in netwons
k = spring constant
x = displacement in metres
equation for time period of a mass oscillating on a spring
T = 2π (√m / k)
T = period of oscillation in seconds
m = mass in kg
k = spring constant
link between time period of the oscillation and the mass of spring
t² ∝ m
link between time period of oscillation of spring and the spring constant
T² ∝ 1 / k
link between time period of oscillation of mass on a spring and amplitude
Time doesn’t depend on amplitude A
equation for elastic strain energy a spring gains from compressing or stretching
E = 1/2 * k * x²
E = elastic energy in joules
k = spring constant
x = displacement in metres
what type of energy do you need to take into account when calculating elastic energy gained for a spring that’s oscillating vertically
graviational potential energy must be taken into account for a spring oscillating vertically
equation for time period of a pendulum
up to what angles of oscillation does this equation apply
T = 2π (√L / g)
T = time period in seconds
L = length of pendulum in metres
g = gravitational field strength Nkg^-1
why will a pendulum clock keep ticking in regular time intervals even if its swing becomes very small
in SHM, the frequency and period are independent of the amplitude
(constant for a given oscillation)
modelling equation for SHM: force (acceleration) and displacement link
since F = ma = m(d²x / dt²)
F = -kx
so:
(d²x / dt²) = -k/m x
equation to work out change in velocity in a fixed time interval
Δv = -(k/m)x Δt
equation for change in displacement compared to v over a fixed time interval
v = Δx / Δt
Δx = vΔt
limitation of simple mathematical model
when run for a long time, it produces an ever increasing amplitude which is not what happens in simple harmonic oscillation, so it’s a limitation of the simple model
equation for velocity (speed) and max velocity
V = +- 2πf (√A² - x²)
Vmax = +- 2πfA
how to work out amplitude of a pendulum from equilibrium if you’re given height diffrerence
you have to get it through energy. at that heigh difference it has no kinetic energy, but it has gravitational potential energy, mgh
when it goes back to equilibrium it has kinetic energy
mgh = 1/2m (Vmax)²
gh = 1/2 (Vmax)²
equation for tension in pendulum’s string
Tension = mg to hold pendulum in place
there’s also circular motion
T = mg + mv² / r
velocity and acceleration of oscillators at any given time with sin and cos
v = sin 2πft
a = cos2πft
features of free oscillations
constant amplitude
no resultant driving force
vibrates at it’s natural frequency
features of forced oscillations
have a driving force, without the driving force the oscillations would slow to a stop
what is resonance
if the driving frequency (frequency of the driving force, an external force) matches the natural frequency of the object / system, resonance occurs.
this causes amplitude of oscillations to rapidly increase
the amplitudes will rise until the energy losses
what happens if a system is only lightly damped and it starts resonating
the increase in aplitude due to resonating will still be dramatic
what is damping
the reverse of resonance.
occurs when energy is lost from an oscillating system, causing amplitude to decrease to 0
usually due to frictional forces like air resistance
levels of dampening for oscillation along with descriptions
lightly damped -
decrease to 0 is gradual, max displacement reduces every oscillation, but the time period is near constant
critically damped -
zero amplitude is reached in the shortest time, oscillator stops at equilibrium without even finishing a cycle
heavily damped (overdamped) -
decrease of displacement is slow and return to equilibrium takes a lot longer
why are systems damped?
to stop them oscillating or to minimise resonance effects
what happens when the driving frequency exceeds the natural frequency
and how does more damping help
the amplitude of the oscillation is even lower than without a driving force
more damping decreases the maximum resonance and there’s a broader maximum peak
equation for energy stored by a SHO
E = 1/2 K A²
equation for max acceleration
at a(max) acceleration is mas as displacement is max so accel = A
a(max) = -4π²f²A
kinetic energy when the osciallator passes through the equilibrium position
at x = 0
all the energy is kinetic energy , Ek
total energy of an oscillator any point
Etotal = 1/2 K A² = 1/2Ek + 1/2Ep
Etotal = 1/2mv² + 1/2kx²
how to find max kinetic energy (total energy) at equilibrium for a swinging pendulum
Kemax = 1/2mvmax²
vmax = 2πfA
so:
Kemax = mπfA
why lines of displacement in an oscillator are curves
kinetic energy is proportional to velocity² (1/2mv²)
and velocity depends on displacement
so kinetic energy is proportional to displacement²
how do you work out total energy of the system
try find out what Vmax is (2πfA)
and then you can plug it into Kemax = 1/2mvmax²
for this simple pendulum the maximum height h above the equilibrium position where the 2 orange balls are is 0.02m
the mass of the pendulum bob is 0.05kg
on the far left is A, on the far right is C, and in the middle is B
describe the changes between GPE and KE as the pendulum bob moves between point A and B
At point A the bob is instantaneously stationary at its maximum height h, so it has 0 kinetic energy and maximum gravitational potential energy
as it goes back towards the position of equilibrium, it is accelerating and so gaining kinetic energy while losing GPE
once it has reached it, kinetic energy is at maximum, while h is 0 so GPE is 0 relative to the point of equilibrium
for a simple pendulum the max height h above equilibrium is 0.02m and the mass of the bob is 0.05kg
calculate the total energy of the system
total energy of the system = maximum potential energy
Etotal = 0.05kg * 9.8N kg^-1 * 0.02m = 9.8 * 10^-3 J
pendulum has heigh h above equilibrium of 0.02m, mass of pendulum bob is 0.05kg
calculate velocity of bob as it passes through equilibrium
at equilibrium, kinetic energy = max and GPE = 0,
so kinetic energy is the total energy, 9.8 * 10^-3 J
1/2mv² = 9.8 * 10^-3 J
v = (√(2 * 9.8 *10^-3) / 0.05kg) = 0.63ms^-1
