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
[tex]v_{max}=\omega A[/tex]
where
[tex]v_{max}[/tex] is the maximum speed
[tex]\omega[/tex] is the angular frequency
A is the amplitude
For the tuning fork in the problem, we have
[tex]\omega=2\pi f=2 \pi(392 Hz)=2462 rad/s[/tex], where f is the frequency
[tex]A=0.600 mm=6\cdot 10^{-4} m[/tex] is the amplitude
Therefore, the maximum speed is
[tex]v_{max}=(2462 rad/s)(6\cdot 10^{-4}m)=1.48 m/s[/tex]
b) [tex]3.0\cdot 10^{-5} J[/tex]
The fly's maximum kinetic energy is given by
[tex]K=\frac{1}{2}mv_{max}^2[/tex]
where
[tex]m=0.0270 g=2.7\cdot 10^{-5} kg[/tex] is the mass of the fly
[tex]v_{max}=1.48 m/s[/tex] is the maximum speed
Substituting into the equation, we find
[tex]K=\frac{1}{2}(2.7\cdot 10^{-5}kg)(1.48 m/s)^2=3.0\cdot 10^{-5} J[/tex]
