In a lecture demonstration, a 3.0-m-long vertical string with ten bolts tied to it at equal intervals is dropped from the ceiling of the lecture hall. The string falls on a tin plate, and the class hears the clink of each bolt as it hits the plate. The sounds will not occur at equal time intervals. How could the bolts be tied so that the clinks occur at equal intervals?

The bolt at the bottom of the string will have a lesser distance to fall than the one at the top. It will therefore arrive at a slower speed. Each successive bolt will arrive a bit faster. The time intervals between equally spaced bolts wil therefore become less each time. Think about how you could make them equal by adjusting the spacing

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aristocles aristocles · Answer 2 · 2019-09-19T03:15:01+02:00

Answer:

they must be arranged in such a way that the distance between two adjacent bolts must be increased in ratio of 1 : 4 : 9 :....

Explanation:

as we know that the time taken by an object to hit the ground from rest position is given as

[tex]h = \frac{1}{2}gt^2[/tex]

[tex]t = \sqrt{\frac{2h}{g}}[/tex]

so here we can say that time to hit the floor directly proportional to the square root of its height from the surface

now we wish to arrange all the bolts in such a way that they all will hit in regular interval of time

so we will have to arrange them such that

[tex]t_1 : t_2 : t_3 ......= \sqrt{h_1} : \sqrt{h_2} : \sqrt{h_3}.......[/tex]

[tex]1 : 2 : 3 : .......... = \sqrt{h_1} : \sqrt{h_2} : \sqrt{h_3} .......[/tex]

so we have

[tex]h_1 : h_2 : h_3 : ...... = 1 : 4 : 9 : .......[/tex]

so they must be arranged in such a way that the distance between two adjacent bolts must be increased in ratio of 1 : 4 : 9 :....