Respuesta :
Answer
given,
frequency of tuning fork = 936 Hz
Amplitude = 0.717 mm
a) The maximum acceleration is given by
[tex]a_{max}= \omega^2 x_{max}[/tex]
[tex]a_{max}= (2\pi f)^2 x_{max}[/tex]
[tex]a_{max}= 4 \pi^2 f^2 x_{max}[/tex]
[tex]a_{max}= 4 \pi^2 936^2 \times 0.717 \times 10^{-3}[/tex]
[tex]a_{max}=2.48 \times 10^4 \ m/s^2[/tex]
b) maximum velocity
[tex]v_{max}= \omega\ x_{max}[/tex]
[tex]v_{max}= 2\pi f y_{max}[/tex]
[tex]v_{max}= 2\pi \times 936 \times 0.717 \times 10^{-3}[/tex]
[tex]v_{max}=4.22\ m/s[/tex]
c) displacement =y = 0.232 mm
[tex]a_{max}= \omega^2 y_{max}[/tex]
[tex]a_{max}= (2\pi f)^2 y_{max}[/tex]
[tex]a_{max}= 4 \pi^2 f^2 y_{max}[/tex]
[tex]a_{max}= 4 \pi^2 936^2 \times 0.232 \times 10^{-3}[/tex]
[tex]a_{max}=8.024 \times 10^3 \ m/s^2[/tex]
d) velocity at y = 0.232 mm
[tex]v_{max}= \omega\sqrt{x_{max}^2-y^2}[/tex]
[tex]a_{max}= 2\pi f\sqrt{x_{max}^2-y^2}[/tex]
[tex]a_{max}= 2\pi \times 936 \times \sqrt{(0.717 \times 10^{-3})^2-(0.232 \times 10^{-3})^2}[/tex]
[tex]a_{max}=3.989\ m/s[/tex]
