Simple Harmonic Motion of a skyscraper

Question

During an earthquake a skyscraper is designed to sway back and forth with a simple harmonic motion with a period of 8 seconds. The amplitude at the top floor for a particular earthquake is 70 cm. 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

Answer 1

During an earthquake a skyscraper is designed to sway back and forth with a simple harmonic motion with a period of 8 seconds. The amplitude at the top floor for a particular earthquake is 70 cm. 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.

Since the period is 8 sec, the frequency, f, is 0.125 sec^-1=0.125 Hz

If the top floor has an amplitude A of 70 cm, then the SHM is:

Asin(2πft)=70sin[2π(0.125)t] cm=70sin(πt/4) cm

b) The speed of the object moving round the circle

What object moving around what circle? If we say π/4 rad/sec is the angular speed, then

V=(π/4 rad/sec)(35 cm)=27.5 cm/sec is the velocity but I'm not sure what all this is asking for.

We can figure the velocity of the top floor by (d/dt)[70sin(πt/4)]=70(π/4)cos(πt/4)=17.5πcos(πt/4) cm/sec

c) The angular velocity. π/4 radians/sec

d) The maximum speed of the top floor 17.5π cm/sec = 55 cm/sec