Tall buildings are flexible and undergo simple harmonic motion (SMH) when excited by wind. Suppose the top floor of a skyscraper is in SMH, swaying back and forth with a frequency of 0.181 Hz and an amplitude of 1.5 m.
a.What is the maximum speed experienced by top floor occupants?
b.What is the maximum acceleration experienced by top floor occupants?
x = 1.5 sin (2 pi * .181 t)
= 1.5 sin (1.137 t)
v = dx/dt = 1.5* 1.137 cos (1.137 t)
v max = 1.5 * 1.137
a = d^2x/dt^2 = -1.5*1.137^2 sin (1.137 t)
a max = 1.5 * 1.137^2
To find the maximum speed experienced by top floor occupants, we can use the formula for maximum speed in simple harmonic motion:
v_max = ω * A
where:
v_max is the maximum speed,
ω is the angular frequency,
and A is the amplitude.
In this case, the frequency of the motion is given as 0.181 Hz. We can convert this to angular frequency using the formula:
ω = 2πf
where:
ω is the angular frequency,
f is the frequency of the motion.
Substituting the given values into the formula, we have:
ω = 2π * 0.181 = 0.362π rad/s
The amplitude is given as 1.5 m.
Now, we can calculate the maximum speed:
v_max = (0.362π rad/s) * (1.5 m) ≈ 1.712 m/s
Therefore, the maximum speed experienced by top floor occupants is approximately 1.712 m/s.
To find the maximum acceleration experienced by top floor occupants, we can use the formula for maximum acceleration in simple harmonic motion:
a_max = ω^2 * A
where:
a_max is the maximum acceleration,
ω is the angular frequency,
and A is the amplitude.
Using the same values as before, we have:
a_max = (0.362π rad/s)^2 * (1.5 m) ≈ 0.623 m/s^2
Therefore, the maximum acceleration experienced by top floor occupants is approximately 0.623 m/s^2.
To find the maximum speed experienced by the top floor occupants, we can use the formula:
v_max = A * ω
where:
- v_max is the maximum velocity or speed,
- A is the amplitude of the motion, and
- ω is the angular frequency.
In this case, the amplitude (A) is given as 1.5 m and the frequency is given as 0.181 Hz. To find the angular frequency, we can use the formula:
ω = 2πf
where:
- ω is the angular frequency, and
- f is the frequency.
Plugging in the values, we have:
ω = 2π * 0.181 Hz
Now we can calculate ω and use it to find v_max:
ω = 2π * 0.181 Hz ≈ 1.14 rad/s
v_max = 1.5 m * 1.14 rad/s
v_max ≈ 1.71 m/s
Therefore, the maximum speed experienced by the top floor occupants is approximately 1.71 m/s.
To find the maximum acceleration experienced by the top floor occupants, we can use the formula for acceleration in simple harmonic motion:
a_max = A * ω^2
where:
- a_max is the maximum acceleration,
- A is the amplitude of the motion, and
- ω is the angular frequency.
Using the given values:
a_max = 1.5 m * (1.14 rad/s)^2
a_max ≈ 2.07 m/s^2
Therefore, the maximum acceleration experienced by the top floor occupants is approximately 2.07 m/s^2.