Respuesta :
A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways?
Answer:
Part a)
[tex]x = 3 mm[/tex]
Part b)
[tex]f = 6.5 Hz[/tex]
Explanation:
Part a)
As we know that shear modulus is given as
[tex]\eta = \frac{F/A}{x/L}[/tex]
now we have
[tex]x = \frac{FL}{\eta A}[/tex]
here we have
[tex]F = 300 N[/tex]
[tex]A = 15 cm^2[/tex]
[tex]L = 3 cm[/tex]
[tex]\eta = 2 \times 10^6 Pa[/tex]
now we have
[tex]x = \frac{300 \times 0.03}{(15 \times 10^{-4})(2\times 10^6)}[/tex]
[tex]x = 3 mm[/tex]
Part b)
As we know that
[tex]F = \frac{\eta A}{L} x[/tex]
now the frequency of oscillation is given as
[tex]f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
here we know
[tex]k = \frac{\eta A}{L}[/tex]
[tex]k = \frac{2\times 10^6 \times (15 \times 10^{-4})}{0.03}[/tex]
[tex]k = 10^5[/tex]
[tex]f = \frac{1}{2\pi}\sqrt{\frac{10^5}{60}}[/tex]
[tex]f = 6.5 Hz[/tex]