Answer:
d = 8.4 cm
Explanation:
In order to calculate the amplitude of oscillation of the top of the building, you use the following formula for the max acceleration of as simple harmonic motion:
[tex]a_{max}=A\omega^2[/tex] (1)
A: amplitude of the oscillation
w: angular speed of the oscillation = 2[tex]\pi[/tex]f
f: frequency = 0.17Hz
The maximum acceleration of the top of the building is a 2.0% of the free-fall acceleration. Then, you have:
[tex]a_{max}=0.02(9.8m/s^2)=0.196\frac{m}{s^2}[/tex]
Then, you solve for A in the equation (1) and replace the values of the parameters:
[tex]A=\frac{a_{max}}{\omega^2}=\frac{a_{max}}{4\p^2i f^2}\\\\A=\frac{0.196m/s^2}{16\pi^2(0.17Hz)^2}\\\\A=0.042m=4.2cm[/tex]
The total distance, side to side, of the oscilation of the top of the building is twice the amplitude A. Then you obtain:
d = 2A = 2(4.2cm) = 8.4cm
The total side to side distance at the top of the building, covered during such an oscillation is 8.4 cm.
Given data:
The height of building is, h = 152 m.
The frequency on windy days is, f = 0.17 Hz.
The acceleration on the top of building is, a = 2/100g (Here g is gravitational acceleration).
This problem can be resolved using the concept of amplitude and angular frequency. The expression for the magnitude of acceleration at the top pf building is given as,
[tex]a = A \times \omega^{2}\\\\a = A \times (2 \pi f)^{2}\\\\\dfrac{2}{100} \times g=A \times (2 \pi \times 0.17)^{2}\\\\\dfrac{2}{100} \times 9.8=A \times (2 \pi \times 0.17)^{2}\\\\A =0.042 \;\rm m[/tex]
And, the total distance, side to side, of the oscillation of the top of the building is twice the amplitude A. Then,
]s = 2A
s = 2 (0.042)
s = 0.084 m
s = 8.4 cm
Thus, we can conclude that the total side to side distance at the top of the building, covered during such an oscillation is 8.4 cm.
Learn more about the oscillatory motion here:
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