A human wave during sporting events within large, densely packed stadiums, spectators will send a wave (or pulse) around the stadium (Figure). As the wave reaches a group of spectators, they stand with a cheer and then sit. At any instant, the width wof the wave is the distance from the leading edge (people are just about to stand) to the trailing edge (people have just sat down). Suppose a human wave travels a distance of 853seats around a stadium in 39 s, with spectators requiring about 1.8 sto respond to the wave’s passage by standing and then sitting. (a)What is the wave speed v(in seats per second) and (b)What is widthw (in number of seats)?

Figure

• A human wave travels a distance in terms of seats, d =853
• The time to travel 853 seats, t =39 s.
• The time required for spectators to respond, t =1.8 s

The speed is equal to the distance traveled by a wave in a given time.

In the given problem, the distance is measured in terms of the number of seats. Therefore, the speed of the wave can be calculated knowing the number of seats the wave has traveled in the given time.

Formula:

Speed of a body in motion, $v=\frac{d}{t}$ (i)

The equation for velocity using equation (i)and the given values is given by:

$$\begin{gathered} v=\frac{853\mathrm{seats}}{39\mathrm{s}} \\ =21.87\mathrm{seats}/\mathrm{s} \\ \cong 22\mathrm{seats}/\mathrm{s} \end{gathered}$$

Hence, the value of wave speed is 22 seats/s

The width can be calculated using the same equation considering the distance as width using equation (i) can be given as:

$$\begin{gathered} w=v\times t \\ =21.87\times 1.8 \\ =39\mathrm{seats} \end{gathered}$$

Hence, the width in terms of seats is 39 seats .