a. An undamped 2.70-kg horizontal spring oscillator has a spring constant of 38.

Question

a. An undamped 2.70-kg horizontal spring oscillator has a spring constant of 38.3 N/m. While oscillating, it is found to have a speed of 3.72 m/s as it passes through its equilibrium position. What is its amplitude of oscillation in meters?

b. What is the oscillator's total mechanical energy as it passes through a position that is 0.722 of the amplitude away from the equilibrium position in J?

c. Ultrasound of intensity 125 W / m2 is produced by the rectangular head of a medical imaging device measuring 2.50 cm by 5.50 cm. What is its power output in W?

Explanation / Answer

a) w = sqrt(k/m) = sqrt(38.3/2.7) = 3.8 rad/sec

v = Aw

A = v/w = 3.72/3.8 = 0.98 m

b) total energy = 1/2kA^2 = 1/2*38.3*0.98^2 = 18.39 J

c) intensity i = p/area

p = 125*0.025*0.055 = 0.172 watt

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