Question

A 1000 kg car carrying four 82 kg people travels over a "washboard” dirt road with corrugations 4.0 m apart. The car bounces with maximum amplitude when its speed is 16 km/h. when the car stops, and the people get out, by how much does the car body rise on its suspension?

• A. 10 cm
• B. 3 cm
• C. 5 cm
• D. None of these

Answer 1

The correct option is C

5 cm

The time period between corrugations would be $T=\frac{4m}{(16\times \frac{5}{18})}\frac{9}{10}s$

$f=\frac{1}{T}=\frac{10}{9};\omega =2\pi f=2\pi \times \frac{10}{9}=\frac{20\pi }{9}$

since at this frequency the amplitude is maximum. So it got to be its natural frequency.

As we know $\omega =\sqrt{\frac{k}{m}}\Rightarrow k={\omega }^{2}m$

$k=(\frac{20{\pi }^2}{9}{)}^2\times (1000+(4\times 82))=64659.5N{m}^{-1}$

when the people get out of the car the weight change is $4\times 82\times 10=3280N$

we have $\omega =kx$

$3280=64659.5x$

$x=0.050m\approx 5cm$