Answer
${\bf 10} \; \rm{kHz}$
Work Step by Step
To find the minimum bandwidth needed to transmit a pulse consisting of 100 cycles of a 1.00 MHz oscillation, we can use the concept of the uncertainty principle for signals.
$$\Delta f \approx \frac{1}{\Delta t}\tag 1$$
Now we need to Calculate the pulse duration $ \Delta t $
We are given that the oscillation frequency is $1.0 \; \rm{MHz} $, and that the pulse consists of 100 cycles, so the total duration of the pulse is:
$$\Delta t = \frac{100}{1.0 \times 10^6} =1\times 10^{-4}\; \rm{s}$$
Plug into (1) to calculate the minimum bandwidth;
$$
\Delta f = \frac{1}{ 10^{-4} } = \color{red}{\bf 10} \; \rm{kHz}
$$
The minimum bandwidth required to transmit the pulse is approximately 10 kHz.
