Question

A tuning fork labeled 392 Hz has the tip of each of its two prongsvibrating with an amplitude of 0.600mma) What is the maximum speed of the tip of a prong?b) A housefly (musca domestica) with mass 0.0270 g isholding on to the tip of one of the prongs. As the prong vibrates,what is the fly's maximum kinetic energy? Assume that the fly'smass has a negligible effect on the frequency of oscillation?

Answer 1

a) 1.48 m/s

The tuning fork is moving by simple harmonic motion: so, the maximum speed of the tip of the prong is related to the frequency and the amplitude by

$v_{max}=\omega A$

where

$v_{max}$ is the maximum speed

$\omega$ is the angular frequency

A is the amplitude

For the tuning fork in the problem, we have

$\omega=2\pi f=2 \pi(392 Hz)=2462 rad/s$, where f is the frequency

$A=0.600 mm=6\cdot 10^{-4} m$ is the amplitude

Therefore, the maximum speed is

$v_{max}=(2462 rad/s)(6\cdot 10^{-4}m)=1.48 m/s$

b)  $3.0\cdot 10^{-5} J$

The fly's maximum kinetic energy is given by

$K=\frac{1}{2}mv_{max}^2$

where

$m=0.0270 g=2.7\cdot 10^{-5} kg$ is the mass of the fly

$v_{max}=1.48 m/s$ is the maximum speed

Substituting into the equation, we find

$K=\frac{1}{2}(2.7\cdot 10^{-5}kg)(1.48 m/s)^2=3.0\cdot 10^{-5} J$