11155299
Description
Flashcards by Noah Bryan, updated more than 1 year ago
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Created by Noah Bryan
over 6 years ago
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| Question | Answer |
| 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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