Question

We can model the motion of a bumblebee's wing as simple harmonic motion. A bee beats its wings 250 times per second, and the wing tip moves at a maximum speed of 2.5 m/s. 

Part A 

What is the amplitude of the wing tip's motion? Express your answer to two significant figures and include the appropriate units. 

Answer 1

Concepts and reason

The concepts used in this problem are angular frequency and the linear speed.

Initially, calculate the angular frequency using the formula of angular frequency. Then, calculate the amplitude using the relation of angular frequency and linear speed.

Fundamentals

Linear speed and angular frequency:

The relation of angular frequency with linear speed is,

$v = \omega A$

Here, $v$ is the linear speed, $\omega$ is the angular frequency and $A$ is the amplitude.

Angular frequency:

The angular frequency measures the angular displacement per unit time. The expression of angular frequency is given as,

$\omega = 2\pi f$

Here, $f$ is the frequency (vibrations per unit time).

The expression of angular frequency is,

$\omega = 2\pi f$

Substitute $250{\rm{ }}{{\rm{s}}^{ - 1}}$ for $f$ in the above expression.

$\begin{array}{c}\\\omega = 2\pi \left( {250{\rm{ }}{{\rm{s}}^{ - 1}}} \right)\\\\ = 1570.8{\rm{ rad/s}}\\\end{array}$

The relation of angular frequency and the linear speed is,

$v = \omega A$

Rearranging the equation for amplitude.

$A = \frac{v}{\omega }$

Substitute $1570.8{\rm{ rad/s}}$ for $\omega$ and $2.5{\rm{ m/s}}$ for $v$ .

$\begin{array}{c}\\A = \frac{{2.5{\rm{ m/s}}}}{{1570.8{\rm{ rad/s}}}}\\\\ = 0.00159{\rm{ m}}\\\\ = 1.59 \times {10^{ - 3}}{\rm{ m}}\\\end{array}$

Ans:

The amplitude of the wing tip’s motion is ${\bf{1}}{\bf{.59 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}{\bf{ m}}$