Simple isn't it? #8

Classical Mechanics Level 5

A group of friends from Brilliant decide to go for a bungee jumping camp. The last one to jump is Rajdeep. He jumps from the top of a cliff of height $H$ .

He performs a damped oscillatory motion along the vertical motion. The first time he reaches at the lowest level i.e. $h_0=0$ m above the ground. He oscillates until he settles at an equilibrium height of $\frac{51}{200}H$ m above the ground.

If the maximum acceleration experienced by Rajdeep during the motion can be written as $\frac{ag}{b}$ , find $a+b$

Details and Assumptions

Nothing bad happens to Rajdeep.
Consider Rajdeep to be a point particle.

This problem is of this set.

The answer is 20.

1 solution

Alex Wang
Jul 19, 2015

This problem can be solved with the same method used to solve the previous problem in this set involving Swapnil.

Let y be the height where the acceleration is maximum.

Then, $mgy = \frac{1}{2}k(y-L)^2$ . We also have $mg = k(\frac{149}{200}-L) = \frac{9}{200}kH$ if you've solved the previous problems in this set.

Substituting for $mg$ into the first equation, we get a quadratic for $\sqrt{y}$ which we can solve to get $y=\frac{49}{100}H.$

Then, $mg - k(y-L) = ma.$ In previous problems, we determined $L = \frac{7}{10} H.$

Finally, we get $a = \frac{17}{3} g$ and our final answer is $\boxed{20}.$

Simple isn't it? #8

The answer is 20.

1 solution

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