During an earthquake a skyscaper is designed to sway back and forth with simple

Question

During an earthquake a skyscaper is designed to sway back and forth with simple harmonic motion with a period of 8 seconds. The amplitude at the top of the flooe for a particular earthquake is 70cm. With respect to the simple harmonic motion of the top floor, calculate the following quantities:

A) the radius of the circle used to represent the SHM

B) The speed of the object moving round the circle

C) the angular Velocity

D) the Maximum speed of the top floor

Please go step by step and provide the formulas!

Explanation / Answer

T = 2pi/w

a) the radius of the circle used to represent the SHM= 70 cm =0.7 m

b) V = Vmax sqroot (1- x^2/ A^2) -----------this will vary depending upon the (x )

c) T he angular Velocityw= 2pi/ T = (,6.28 )/ (8 ) =0.785 rad/sec

D) the Maximum speed of the top floor v = wr = (0.785 rad/sec) (0.7m ) =0.5495 m/s

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