Ocr m3: Shm

This is a question I got stuck on in the OCR M3 Textbook p.75.

12. A bungee jumper falls a total distance of 48m, of which the first 20m are free fall under gravity alone before the rope becomes taut. Find the total time she takes to fall to the lowest point of her jump, and the greatest spped she attains.

I found "x" (acceleration) = -9.8(x/28-1), but got stuck from this point. I also found the time to fall under gravity as 2.02 seconds.

Any help would be appreciated.

From energy conservation $48mg = \frac{\lambda \times 28^2}{2l} \Rightarrow 392 \lambda = 48mgl$

useing $\ddot{x} = \frac{mg - T}{m} = g(1- \frac{6x}{49})$
you can see that the acceleration is 0 when $x = \frac{49}{6}$

so the amplitude is $a = 28- \frac{49}{6} = \frac{119}{6}$

useing $\frac{1}{2} T = \frac{\pi}{\omega} = \frac{7 \pi}{\sqrt{6g}}$
and useing $v=\omega a$

you should be able to find the max speed and the rest from here

Also you get the wrong answer for the maximum speed using your values of omega and amplitude. You get 119/6 * root(6)*g/7, when you should get 21.7.