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A car with a total mass of 1190 kg (including passengers) is driving down a washboard road with bumps spaced 5.0 m apart. The ride is roughestβ€”that is, the car bounces up and down with the maximum amplitudeβ€”when the car is traveling at 5.3 m/s. The maximum net force of the car is 1.5*10^3 what is the maximum amplitude of the oscillation?

Respuesta :

Answer:

x = 2.84 cm

Explanation:

First we can find the frequency of the oscillation given as

[tex]f = \frac{v}{d}[/tex]

here we know that

d = 5.0 m

v = 5.3 m/s

now we have

[tex]f = \frac{5.3}{5}[/tex]

[tex]f = 1.06 Hz[/tex]

now we know that

[tex]f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}[/tex]

so we have

[tex]1.06 = \frac{1}{2\pi} \sqrt{\frac{k}{1190}}[/tex]

[tex]k = 5.28 \times 10^4 N/m[/tex]

Now by the formula of spring force we know that

[tex]F = kx[/tex]

so we have

[tex]1.5 \times 10^3 = (5.28 \times 10^4) x[/tex]

[tex]x = 0.0284 m[/tex]

[tex]x = 2.84 cm[/tex]

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