Showing posts with label esercizi oscillazioni. Show all posts
Showing posts with label esercizi oscillazioni. Show all posts

Monday 14 May 2012

Pendulum

A simple pendulum is 5.00 m long. (a) What is the period of simple harmonic motion for this pendulum if it is located in an elevator accelerating upward at 5.00 m/s2? (b) What is its period if the elevator is accelerating downward at 5.00 m/s2? (c) What is the period of simple harmonic motion for the pendulum if it is placed in a truck that is accelerating horizontally at 5.00 m/s2
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is pulled horizontally so that it stretches the spring 5.00 cm and is then released from rest at t = 0. (a) What is the force constant of the spring? (b) What are the angular frequency ω, the frequency, and the period of the motion? (c) What is the total energy of the system? (d) What is the amplitude of the motion? (e) What are the maximum velocity and the maximum acceleration of the particle? (f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. 

Tires

While riding behind a car traveling at 3.00 m/s, you notice that one of the car’s tires has a small hemispherical bump on its rim, as in Figure. (a) Explain why the bump, from your viewpoint behind the car, executes simple harmonic motion. (b) If the radius of the car’s tires is 0.30 m, what is the bump’s period of oscillation?

Saturday 26 March 2011

Oscillazione smorzata

Consideriamo una oscillazione smorzata. Vediamo come varia la funzione al variare del coefficiente di smorzamento, confrontandola con la pura oscillazione.



Il primo massimo (vedi figura) si sposta verso t=0 al crescere del fattore di smorzamento B. Le curve rosse sono a diversi valori crescenti di smorzamento. La curva verde è la funzione trigonometrica per A=1. Nella figura, le curve smorzate hanno B=omega/4.*i, con i da 1 a 8.
Calcoliamo la derivata per trovare il massimo



Il primo massimo (vedi figura) si sposta verso t=0 al crescere del fattore di smorzamento B. Si trova t risolvendo la funzione tan(omega*t)=omega/B.
Oppure lo si trova sul grafico.