PH10053 - Sergey's Theory

Descripción

Simple Harmonic Motion- Damped Oscillations - Harmonics
Noah Bryan
Fichas por Noah Bryan, actualizado hace más de 1 año
Noah Bryan
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Resumen del Recurso

Pregunta Respuesta
Define Periodic Motion Motion of an object that regularly repeats - An the object returns to given position after a fixed time interval - A feature of Simple Harmonic Motion
Name the Force that acts to reduce displacement of an object and return the system to equilibrium. Restoring Force
SHM occurs when... The Restoring Force is linearly proportional to the displacement from from equilibrium and opposite to the direction of the displacement.
A general rule for SHM is: Restoring Force=-k(x)
Name two assumptions: 1) No Friction 2) Mass of spring is negligibly small in comparison to the mass of the block.
When an object is displaced, the spring exerts a ...... ...... equal to .... Restoring Force ; (-kx) where k is the force constant.
What's the differential equation for simple harmonic motion? What's its general solution? (d^2x/dt^2)+(w^2)x=0 x=Acos(wt+d)
Quantities: Period (T) Frequency (f) Angular Frequency (w) Equations and Units: Period (s) Frequency --> f=1/T (Hz) Angular Frequency --> w=2πf (rad/s)
Differentiating the general solution: x=Acos(wt+d) gives: v=... and a=... v=-Awsin(wt+d) a=-A(w^2)cos(wt+d) = -(w^2)x
Phase difference of Velocity and Displacement π/2
Velocity is maximum when... Displacement is 0 (Eqm position)
Velocity is 0 when... Displacement is maximum (Amplitude of oscillation)
The frequency and period are related to the stiffness (k) of the spring and the particle mass; where w= sqrt(k/m) f=w/2π=(sqrt(k/m))/2π T=1/f= 2π(sqrt(m/k))
When is the spring constant not constant? When oscillations are large
What's the equation for the vertical displacement in equilibrium when a mass is attached to a spring? y.=mg/k
y' for vertical oscillations is the... Amplitude of oscillations about the equilibrium position of the system when the mass is attached. Where y'=y-y., where y is the total amplitude from the eqm position without the mass attached
Eqm Position: y.=mg/k Hence, mg-ky.=0 Write the equation for vertical oscillations: mg-ky.=ma by Newton's (II) Law This translates to: m((d^2)y/d(t^2))+ky.=mg (x(1/m)) --> ((d^2)y/d(t^2))+ (k/m)y.=g; where (k/m) = w
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