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Four springs have been compressed from their equilibrium position at x = 0 cm. When released, they will start to oscillate. Rank in order, from highest.

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Presentation on theme: "Four springs have been compressed from their equilibrium position at x = 0 cm. When released, they will start to oscillate. Rank in order, from highest."— Presentation transcript:

1 Four springs have been compressed from their equilibrium position at x = 0 cm. When released, they will start to oscillate. Rank in order, from highest to lowest, the maximum speeds of the oscillators. c > b > a = d c > b > a > d d > a > b > c a = d > b > c b > c > a = d STT14.3

2 Four springs have been compressed from their equilibrium position at x = 0 cm. When released, they will start to oscillate. Rank in order, from highest to lowest, the maximum speeds of the oscillators. c > b > a = d c > b > a > d d > a > b > c a = d > b > c b > c > a = d STT14.3 The potential energy stored in the springs gets converted to kinetic energy of the ball at the equilibrium location. ½kx2 = ½mv2, so |v| = |x|*sqrt(k/m).

3 Velocity is positive; force is to the right.
This is the position graph of a mass on a spring. What can you say about the velocity and the force at the instant indicated by the dotted line? Velocity is positive; force is to the right. Velocity is negative; force is to the left. Velocity is negative; force is to the right. Velocity is zero; force is to the right. Velocity is zero; force is to the left. STT14.4

4 Velocity is positive; force is to the right.
This is the position graph of a mass on a spring. What can you say about the velocity and the force at the instant indicated by the dotted line? Velocity is positive; force is to the right. Velocity is negative; force is to the left. Velocity is negative; force is to the right. Velocity is zero; force is to the right. Velocity is zero; force is to the left. STT14.4 The slope of the position graph (or first derivative) gives the velocity. The slope of the graph of the slope of the position graph (or second derivative) gives the acceleration. The slope is increasing from negative to positive in the neighborhood of the point in question, so the acceleration is positive. The sign of the force matches the sign of the acceleration by Newton’s second law.


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