oscillations Flashcards
define simple harmonic motion
it is the motion of an object from a fixed point where its acceleration is proportional to its displacement and it always acts in the direction towards that fixed point
formula for acceleration in shm
a = -w^2 * x
where w = angular frequency
x = displacement
what is the condition for a particle in shm
acceleration is directly proportional to -x, where w is a constant
define angular frequency
the rate of change of phase angle of the oscillation, and is equal to the product of 2pie and its frequency
define amplitude
the magnitude of the maximum displacement of the particle from its equilibrium position
define frequency
the number of oscillations per unit time
define period
the time it takes to complete one oscillation
explains what happens to velocity as it approaches x0 (i.e. max displacement)
when approaching max point, velocity always decreases as it approaches the amplitude, because acceleration is always acting at the opposite direction
what happens to velocity at the equilibrium position
velocity is a maximum at the equilibrium position
what happens to acceleration as it approaches x0
acceleration is a maximum at x0 and is always pointing towards the direction of the equilibrium point
state the equations of x, v, and a as a function of time when x=x0 at t=0
x = x0coswt
where:
x0 = amplitude,
w= angular frequency
t= time
v= -wx0sinwt
a=-w^2x0coswt
state the equations of a and v as a function of x
v= +-w * sqr root of (x0^2 - x^2)
a= -w^2*x
define free oscillation
it occurs when an object oscillates with no resistive and driving forces acting on it. Its total energy and amplitude remain the same with time
in a spring mass system, horizontal or vertical, what are the factors that affect the objects motion
only mass and spring constant
how to find out:
for horizontal:
F restoring = -kx
by newton’s second law
ma=-kx
a= -(k/m)x
comparing a= -w^2 * x
w^2 = k/m
w= sqr root k/m
T= 2pie * sqr root m/k
for vertical:
at equilibrium
mg=ke -(1)
once block is pulled down, for vertical components,
Fnet = F restoring = W + F spring = mg + [-k(e +y)]
where e represents extension of string when mass placed while y = extension of string from equilibrium point (obtained after mass is placed so does not include e) after mass is pulled and released from that point
by newton’s second law
ma = mg -ke - ky
from -(1)
ma = ke - ke - ky
ma = -ky
a= -(k/m)y
from a= -w^2 * y, where y is the displacement from the equilibrium
w= sqr root k/m
T = 2pie * sqr root m/k
what are the factors that affect an objects motion in a pendulum during free oscillations
only depends on acceleration due to gravity and length of the pendulum
proving:
u have to use small angle approx for this:
take note that the restoring force is ALWAYS TANGENTIAL TO THE OBJECTS MOTION, therefore we only consider the weight of the object and not the tensional force as tensional force is perpendicular to the path of the object so it does not contribute to the restoring force
F restoring= -mgsintheta
since theta is very small, by small angle approx:
sin theta = theta
F restoring = -mgsintheta = -mgtheta =-mg(s/L) , as theta = s/L , where s is arc length, aka displacement from equilibrium point, and L is radiues
by newton’s second law,
ma=-mg(s/L)
a= -(g/L)s
comparing a with -w^2 * s
w= sqr root (g/L)
T= 2pie * sqr root (L/g)
