Physics: Oscillations, Wave Motions, Superposition

sign convention for SHM

follow displacement, opposite to direction of force

ω =

2πf or 2π/T

how to find max speed?

max speed occurs at eqb. By POCOE, loss in epe = gain in ke. x for epe = max displacement.

equations of displacement --> vel --> acc

usually x = x0*sin(ωt)

however, if it is released from max disp, then x = x0*cos(ωt)

for v, differentiate x, for a, differentiate v

graphical relation of x,v,a curves

if x = +sin graph, v = +cos graph, a = -sin graph

if x = +cos graph, v = -sin graph, a = -cos graph (a ∝ -x)

amplitude of x,v,a curves

A = x0

x: A

v: ωA

a: ω^2*A

velocity-displacement graph

v0 = ωx0

Acceleration-displacement graph

A = x0

why is bouncing ball not SHM

(in all prove SHM qsn, prove that a = -ω^2x)

ball is alw subjected to force of gravity downwards when its in motion.

acc is alw downwards and is constant, no eqb position

thus ball does not satisfy a ∝ -x, not SHM

SHM qsn

1. take sign convention to follow disp

2. by N2L, Fnet = ma (then write the eqn for Fnet which will be negative value)

3. you will arrive at a = -Cx (C = ω^2)

4. "thus motion is simple harmonic where ω^2 = C"

does the period of a damped oscillation increase with time

no. T is constant in oscillation

is the frequency of damped osc < undamped?

yes. natural frequency is higher during free osc

graphical significance of a frequency-response graph

Y = amplitude of the oscillating system. X = frequency. C = driver amplitude. its NEVER zero. the peak is when Fd = Fn. max amp, max transfer of energy. when the system has zero damping, the Fn becomes a vertical asymptote. a more damped system will have a lower amplitude and peak at a lower frequency

resonance

resonance is a phenomenon in which an oscillating system responds with maximum amplitude to an external periodic driving force when the frequency of the driving force equals to the natural frequency of the system. At resonance, there is a maximum transfer of energy from the driving system to the driven system.

significance of damping in vel-disp graph

spiral inwards clockwise. start from either x0 or -x0 as mentioned in the question

both velocity and displacement amplitudes shrink toward zero due to energy loss.

energy formulae in shm

find eqn of v. KE = 1/2 m v^2

TE = KEmax = 1/2 m (ωx0)^2

PE = TE - KE

restoring force

net force that restores system back to equilibrium

to sketch disp-time graph from disp-dist graph

shift the curve a little in the direction of propogation of wave

are sound waves polarised?

sound waves are longitudinal - they cannot be polarised

phase difference formula

ϕ = ∆x/λ *2π = ∆t/T *2π

v=

fλ

when a wave enters a new medium, what remains unchanged: wavelength, speed, frequency

frequency

intensity =

k(Amplitude)^2 = power/area = rate of energy transfer per unit area across a surface perpendicular to the direction of wave propagation

State how a polarized transverse wave differs from an unpolarised transverse wave.

polarised: oscillations of the particles are confined to a single plane. no restriction for unpolarised

area for intensity from point source

A = 4πr^2, thus I ∝ 1/r^2

what happens to I and A when passing through a polariser

I is halved. A does NOT change.

malus' law

I = I₀ cos²θ

identify R and C from disp-time curve

+ve gradient = R, -ve gradient = C

deduce D/C or I/C

phase diff due to source + phase diff due to path difference = n(pi)

if odd integer of pi, waves meet in antiphase, interfere destructively. vice versa

how are stationary waves formed?

Stationary waves are formed when two waves of the equal amplitude and frequency travelling with the same speed but in opposite directions superpose

when both ends are closed

• nodes at both ends

• fn = nv/2L, n = 1,2,3… (all harmonics are possible)

when both ends are open

• antinodes at both ends

• fn = nv/2L, n = 1,2,3… (all harmonics are possible)

one open end, one closed end

• node at closed end

• antinode at open end

• fn = nv/4L, n = 1,3,5… (only odd harmonics possible)

what happens to the signal in a stationary radio wave?

nodes - small / zero amplitude --> weak signal - static sound

antinodes - large / max amplitude --> strong signal - clear audio

double slit formula

x = λD/a

x = distance between two bright fringes or two dark fringes

D = dist bw slit and screen

a = dist bw the two slits

central maxima is 0th order.

single slit formula

sin(θmin) = λ/b = dist between central fringe and first minima

θmin = angle between central maxima and first minima

b = width of the slit

note: central fringe = central maxima, because the central fringe is a maxima (bright)

central fringe is brighter + twice as wide as the other fringes

rayleigh's criterion

two sources are just resolved if the central maximum of one diffraction pattern falls on the first minimum of the diffraction pattern of the other

rayleigh criterion formula

θ = λ/b

θ = angle between the two sources with the vertex being the middle of the slit.

b = width of slit

(for my info: from sin(θ) = λ/b, since θ is small, sinθ = θ (mf27)