SHM and Circular Motion Problem

[acomet]

Q: During an earthquake, a skyscraper is designed to sway back and forth with simple harmonic motion with a period of 8 secs. The amplitude at the top floor of 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 at the top floor

I am in particular confused about part a). Would the solving of this question require the use of circular motion equations as well, or would SHM-related equations be enough to answer this?

 

[Simon Bridge]

I am in particular confused about part a). Would the solving of this question require the use of circular motion equations as well, or would SHM-related equations be enough to answer this?

Neither - it does not require any equations at all - only geometry.

You should not be doing these problems by trying to remember which equation goes where.
How does a circle represent SHM?

 

[BiGyElLoWhAt]

Can you express the motion of the top floor as a function of time? I think if you wrote it out, It might be more obvious what you're looking for...

*HINT* Think unit circle...

 

[acomet]

So would the radius of the circle be equal to the amplitude?

 

[HalfLostAndSearching]

I can provide a response to this content by explaining the relationship between simple harmonic motion (SHM) and circular motion and how it applies to the given problem.

To start, let's define SHM and circular motion. SHM is a type of periodic motion in which the restoring force is directly proportional to the displacement from the equilibrium point. This means that the object will oscillate back and forth around the equilibrium point with a constant period. Circular motion, on the other hand, is the movement of an object along a circular path at a constant speed.

Now, in order to solve the given problem, we need to understand that SHM and circular motion are closely related. In fact, SHM can be represented as a projection of circular motion onto a straight line. This means that the motion of the top floor of the skyscraper can be represented as a point moving along a circular path.

a) To find the radius of the circle used to represent the SHM, we can use the formula for the period of SHM, T = 2π√(m/k), where m is the mass of the object and k is the spring constant. Since the period is given as 8 seconds, we can rearrange the equation to solve for the radius, r. This gives us r = T^2k/4π^2m. Therefore, the radius of the circle used to represent the SHM is dependent on the mass and spring constant of the object.

b) The speed of the object moving round the circle can be calculated using the formula v = ωr, where ω is the angular velocity and r is the radius. Since we have already calculated the radius in part a), we just need to find the angular velocity. This can be done using the formula ω = 2π/T, where T is the period. Substituting the given period of 8 seconds, we get ω = π/4 rad/s. Therefore, the speed of the object moving round the circle is v = (π/4)(0.7) = 0.55 m/s.

c) The angular velocity can be calculated using the formula ω = 2π/T, where T is the period. Substituting the given period of 8 seconds, we get ω = π/4 rad/s.

d) The maximum speed at the top floor can be calculated using the formula v = ωA