SHM Flashcards
Defining relationship
a= -(ω^2)x
How to check for shm
- Check for fixed equilibrium position
- Check if acceleration is directly proportional to displacement, and in the opposite direction
Supplementary derivation of x equation
x = Acos(ωt+ϕ)
Solution to x’’=-(ω^2)x
(x’’+(ω^2)x = 0)
A is amplitude
ϕ is original displacement (t=0)
T = 2π/ω (Period doesn’t depend on amplitude or phase)
Types of damping
Light - T is independent of amplitude, so each cycle takes the same amount of time as oscillations decrease in amplitude (sin graph modulated onto exp decay)
Critical - Enough to stop the system oscillating after being released from equilibrium (in shortest time possible without overshooting). Important in mass spring systems e.g. car suspension systems
Heavy - Damping so strong that displaced object returns to equilibrium much more slowly than if it was critically damped - no oscillating motion, e.g. a mass on a spring in thick oil
Natural frequency
When a system oscillates without a periodic force being added
SHM of water in u shaped tube
https://www.youtube.com/watch?v=L9M7Tq90bco
Forced vibrations
When a periodic force is applied to an oscillating system
Effect of increasing applied frequency
Amplitude increases until a max at a particular frequency (resonant frequency), then begins to decrease again
Phase difference between applied force and displacement increases from 0 toπ/2 (max amplitude),
Resonance
When system is oscillating at maximum amplitude, phase diff between displacement and restoring force is π/2 - periodic force is therefore in phase with velocity
Lighter the damping - larger amplitude is at resonant frequency, + closer resonant frequency is to the natural frequency of the system (sharper resonance curve)
As applied frequency increases, amplitude becomes smaller and smaller. For systems with little/no damping, resonant frequency = natural frequency
Resonance occurs at slightly lower frequency - lighter the damping, closer it is to natural. At resonance, periodic force acts at same point in each cycle, increasing amplitude to a maximum
m in shm
Mass of entire oscillating body, e.g. in u tube, m is mass of whole water, in the case of prongs, entire mass. So distance
Conditions for millenium bridge hazard + precautions engineers would take
- A structure has a natural frequency (or
frequencies) of vibration. - Resonance.
- This occurs when the frequency of a driving
force is equal to a natural frequency of the
structure. - Large amplitude vibrations are then produced
(or there is a large transfer of energy to the
structure). - This could damage the structure (or cause a
bridge to fail) - Install dampers (shock absorbers).
- Stiffen (or reinforce) the structure.
- Any other acceptable step e.g. redesign to
change natural frequency, increase the mass of
the bridge, restrict numbers of pedestrians
How to change spring constant
Change unstretched length. If u cut it in half, K doubles
If the spring is cut into half and the same force F is applied to the half spring, the distance it should stretch is X/2, half of the original deformation. From this the new spring constant, K2 becomes K2= F/(X/2) or 2* (F/X) which is double of K.
To understand why this happens, we have to consider the way force acts on a spring. Imagine a spring hanging from a beam with n parts to it. At the nth part a force G is applied. This force acts downward on the nth part. However, to make sure that the nth part doesn’t keep accelerationapproaches downward, a resultant force of the same magnitude as G is applied by the (n-1)th part on the nth part to keep it from falling. Since, the n-1th part is pulling the nth part up, by Newton’s third law, n is pulling n-1 down with force G as well. To offset this force downwards on n-1, n-2 has to pull n-1 up with a magnitude of G as well, and so on and so fort until you reach the top where the 1st part is being pulled down by the 2nd part, while at the same time being pulled up by an immovable beam, that is being pulled down by the 1st part with magnitude G.
Notes for frequencies around resonance (“phase lag”)
f<natural f - Amplitude similar to that of the driver
almost in phase w system
= natural f - amplitude greater than natural A, 90 outta phase
Greater - phase lag of almost 180. Amplitude is v small
Spring P w a ball oscillating in water vs spring Q w a stiffer spring oscillating in honey what diff between graphs of f against A
Effect due to the stiffness of the spring:
Spring Q is stiffer than spring P. This means that it takes more force to stretch or compress spring Q than it does for spring P.
A stiffer spring has a higher resonant frequency than a less stiff spring. This means that it vibrates more rapidly when excited by an external force.
The peak associated with spring Q on the amplitude-frequency graph will therefore be shifted to the right. This is because the resonant frequency of the system is higher for spring Q than for spring P.
Effect due to the viscosity of the liquid:
Honey is more viscous than water. This means that it resists flow more than water does.
In honey, the mass will experience more damping than it would in water. This is because the viscosity of honey leads to greater resistance to motion and therefore greater damping.
Greater damping leads to a reduction in the amplitude of the oscillation, which in turn leads to a lower peak in the amplitude-frequency graph.
Therefore, the peak associated with spring Q will be lower than that of spring P on the amplitude-frequency graph. This is because the oscillation of spring Q in honey is more damped than the oscillation of spring P in water.
Graph of pendulum oscillating in air vs a vaccum
When the pendulum is oscillating in air, there is damping due to air resistance. Damping affects the system in two ways: it causes the resonant frequency to decrease and the amplitude to decrease at the resonant frequency.
The decrease in resonant frequency is due to the energy loss experienced by the pendulum as it oscillates. This energy loss causes the system to resonate at a lower frequency, which is represented by the shift to the left of f0 on the graph.
The decrease in amplitude at the resonant frequency is due to the energy being spread over a broader range of frequencies because of damping. As a result, the graph becomes wider from the top, indicating that the amplitude has decreased
