A theatre spectator and a radio listener hear the opening…

A theatre spectator and a radio listener hear the opening sound of an overture. (a) At what distance from the orchestra should the spectator sit so that he hears the first sound at the same time as a radio listener who is $d = 7500\ \mathrm{km}$ away from the theatre? (b) At what distance from the receiver should the radio listener be so that he hears the first sound at the same time as a spectator sitting $l_1 = 30\ \mathrm{m}$ from the orchestra? The speed of sound is $c = 340\ \mathrm{m/s}$ and the speed of radio waves (light) is $c_d = 3 \cdot 10^8\ \mathrm{m/s}$.

Explanation of Method

Hearing an overture in the theatre and on the radio.

Step 1: How the sound reaches each person

Spectator: the sound travels through the air of the hall at $c = 340\ \mathrm{m/s}$.

Radio listener: a microphone right next to the orchestra picks up the sound, and radio waves carry the signal the whole 7500 km at $c_d$. Only at the end, from the loudspeaker of the receiver to the listener's ear, does it travel as sound again.

The radio delay over 7500 km is

\[ t_r = \frac{d}{c_d} = \frac{7.5 \cdot 10^6\ \mathrm{m}}{3 \cdot 10^8\ \mathrm{m/s}} = 0.025\ \mathrm{s}. \]

$\text{(a)}$ $\text{(b)}$ $\text{spectator}$ $\text{radio listener}$ $\text{spectator}$ $\text{radio listener}$ $\text{sound, } 8.5\ \mathrm{m}$ $\text{radio, } 7500\ \mathrm{km}$ $\text{sound, } 30\ \mathrm{m}$ $\text{radio}$ $\text{sound, } 21.5\ \mathrm{m}$ $0$ $0.025\ \mathrm{s}$ $0.088\ \mathrm{s}$

sound in air (340 m/s)
radio waves (3·10⁸ m/s)

Time runs to the right. In each part both people must hear the first note at the same moment (dashed line). Radio covers 7500 km in 0.025 s, the same time sound needs for only 8.5 m.

Step 2: (a) Where the spectator must sit

Assume the radio listener sits right next to the receiver. The spectator must hear the sound after the same 0.025 s: \[\frac{x}{c} = t_r \;\Longrightarrow\; x = c\,t_r = 340\ \mathrm{m/s} \cdot 0.025\ \mathrm{s}.\]

\[x = 8.5\ \mathrm{m}\]

Step 3: (b) Where the radio listener must sit

The spectator at $l_1 = 30\ \mathrm{m}$ hears the sound after

\[ t_1 = \frac{l_1}{c} = \frac{30\ \mathrm{m}}{340\ \mathrm{m/s}} \approx 0.088\ \mathrm{s}. \]

The radio listener (still 7500 km away) sits a distance $y$ from his receiver. His total delay is the radio delay plus the sound delay in his room: \[t_r + \frac{y}{c} = \frac{l_1}{c} \;\Longrightarrow\; y = l_1 - c\,t_r = 30\ \mathrm{m} - 8.5\ \mathrm{m}.\]

\[y = 21.5\ \mathrm{m}\]

Check

(a) $ \quad \dfrac{8.5\ \mathrm{m}}{340\ \mathrm{m/s}} = 0.025\ \mathrm{s} = t_r\ \checkmark $

(b) $ \quad 0.025\ \mathrm{s} + \dfrac{21.5\ \mathrm{m}}{340\ \mathrm{m/s}} = 0.025 + 0.063 = 0.088\ \mathrm{s} = \dfrac{30}{340}\ \mathrm{s}\ \checkmark $

Result

\[\boxed{\,x = 8.5\ \mathrm{m}\,}\]

\[\boxed{\,y = 21.5\ \mathrm{m}\,}\]

Remark

About the original solution. It let the sound travel the 7500 km through the air, which gave $x = 7.5 \cdot 10^6\ \mathrm{m}$ in (a) and 30 m in (b). But the long distance is covered by radio waves at the speed of light; only a few metres are covered by sound. That is why the speed $c_d$ is given. If in (b) the radio listener were so close to the theatre that the radio delay could be ignored, the answer would be $y = l_1 = 30\ \mathrm{m}$.