Chapter 2, Problem 16Q

To determine

To explain: The reason for occurance of sound at unequal time intervals. Whether the clinking of time increase or decrease near the end of the fall. Also, the method to tied the bolts for the occurance of clinks at equal intervals.

Explanation of Solution

Introduction:

From Newton’s second law of motion,

  $s=ut+\frac{1}{2}a{t}^2$

Each bolt is tied at an equal distance to each other. Initially, bolts are at rest position. The higher the bolt from the ground, the greater distance it would have to travel with greater speed. Hence, less time interval, it will have to travel.

As the distance between bolts is increasing uniformly about the ground. The higher the bolt will be tied; it will have to travel a greater distance to reach the ground. Now, gravity is acting equally on all the bolts. Hence, the higher the distance of the fall, the greater the speed they will acquire in their travel. Even if they are tied at equal distance, bolts will reach the ground in lesser time, and the sound of clink will be heard frequently.

Here, acceleration due to gravity will be applied to the bolts equally. While falling on the ground, positive acceleration will be applied to it.

Thus,

  $\begin{array}{l}s=\frac{1}{2}g{t}^2\\ s\propto t^2\end{array}$

Now, to hear clink sound at the equal interval, their distance would be varying as per the above formula.

Thus, time is such as

  $\begin{array}{l}d{t}_1=1\sec \\ d{t}_2=1\sec \\ t_2=2\sec \end{array}$

The distance between bolts would be such as $1:4:9:16:\mathrm{......}$ .

Conclusion:

Thus, the sound will be heard at unequal time intervals. Because the higher the velocity it will acquire from a height, the sooner they will hit the ground and produce the sound of the clink. Thus, the time between each clink will decrease successively near the end of the fall. Thus, to hear clink sound at equal time intervals, the distance between each bolt would be such as $1:4:9:16:25:36:49:64:81:100$ .