The source is stationary and the detector moves towards it. The Doppler shift formula is: \[ \nu = \nu_0 \left( \frac{c + v_D}{c} \right) \] Given \(\nu = 275\) Hz, \(\nu_0 = 262\) Hz, \(c = 330\) m/s. \[ \frac{275}{262} = 1 + \frac{v_D}{330} \implies v_D = 330 \left( \frac{275}{262} – 1 \right) = 330 \left( \frac{13}{262} \right) \approx 16.37 \text{ m/s} \]

The detector hears the beep over a distance \(\Delta x = 78\) m. The duration of the reception \(\Delta t_{rec}\) is the time it takes the detector to cover this distance: \[ \Delta t_{rec} = \frac{\Delta x}{v_D} \] The number of wave cycles \(N\) emitted is equal to the number of cycles received. \(N = \nu_0 \Delta t_{emit} = \nu \Delta t_{rec}\). Therefore, the duration of the emitted beep \(\Delta t_{emit}\) is: \[ \Delta t_{emit} = \frac{\nu}{\nu_0} \Delta t_{rec} = \frac{\nu}{\nu_0} \frac{\Delta x}{v_D} \]

Substituting the expression for \(v_D = c \frac{\nu – \nu_0}{\nu_0}\): \[ \Delta t_{emit} = \frac{\nu}{\nu_0} \frac{\Delta x}{c (\frac{\nu – \nu_0}{\nu_0})} = \frac{\nu \Delta x}{c (\nu – \nu_0)} \] Plugging in numbers: \[ \Delta t_{emit} = \frac{275 \times 78}{330 \times (275 – 262)} = \frac{21450}{330 \times 13} = \frac{21450}{4290} = 5.0 \text{ s} \]