Question

The wings of the blue-throated hummingbird, which inhabits Mexico and the southwestern United States, beat at a rate of up to 900 times per minute.

Required:
a. Calculate the period of vibration of the bird’s wings.
b. Calculate the frequency of the wings’ vibratio
c. Calculate the angular frequency of the bird’s wingbeats.

Answer 1

a) the period of vibration of the bird’s wings is 0.0667s

b) the frequency of the wings’ vibration is 15 Hz

c) the angular frequency of the bird’s wingbeats is 94.2 rad/s

Step-by-step explanation:

Given the data in the question;

beat at a rate of up to 900 times per minute.

The number of vibrations per unit time can be be determined as follows;

frequency f = number of vibration / time ---------- let this be equation 1

Also Time period T is;

T = 1/f { time period and frequency are reciprocal to each other }

T = 1/f ---------- Let this be equation 2

Then, the angular frequency ω can be determined using the formular;

ω = 2π × f -------------- Let this be equation 3.

Now, given that; The wings of the blue-throated hummingbird beat at a rate of up to 900 times per minute; Hence,

from equation 1, frequency f = number of vibration / time

f = 900 / 1 min

f = 900 / 60 sec

f = 15 Hz

Therefore, the frequency of the wings’ vibration is 15 Hz

a) the period of vibration of the bird’s wings

from equation 2, T = 1/f ,

we substitute in the value of f

T = 1 / 15 Hz

T = 0.0667s

Therefore, the period of vibration of the bird’s wings is 0.0667s

c) the angular frequency of the bird’s wingbeats;

from equation 3, ω = 2π × f

we substitute

ω = 2π × 15 Hz

ω = 94.2 rad/s

Therefore, the angular frequency of the bird’s wingbeats is 94.2 rad/s