If the periodic time of a body $$ (\mathrm{X}) $$ is double that of a body $$ (\mathrm{Y}) $$ so the ratio between th equency (Y) to the frequency $$ (\mathrm{X}) $$ is $$ \ldots \ldots $$

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Given that frequency $$ f = \frac{1}{T} $$ and the periodic time of body $$ (\mathrm{X}) $$ is double that of body $$ (\mathrm{Y}) $$, we have:

$$
T_{X} = 2T_{Y}
$$

Now, the frequency for each body is:

$$
f_{X} = \frac{1}{T_{X}} \quad \text{and} \quad f_{Y} = \frac{1}{T_{Y}}
$$

To find the ratio of $$ f_{Y} $$ to $$ f_{X} $$:

$$
\frac{f_{Y}}{f_{X}} = \frac{\frac{1}{T_{Y}}}{\frac{1}{T_{X}}} = \frac{T_{X}}{T_{Y}}
$$

Substitute $$ T_{X} = 2T_{Y} $$ into the equation:

$$
\frac{f_{Y}}{f_{X}} = \frac{2T_{Y}}{T_{Y}} = 2
$$

Thus, the ratio between the frequency of body $$ (\mathrm{Y}) $$ and the frequency of body $$ (\mathrm{X}) $$ is:

$$
\frac{f_{Y}}{f_{X}} = 2
$$
The frequency of body $$ (\mathrm{Y}) $$ is twice that of body $$ (\mathrm{X}) $$, so the ratio $$ \frac{f_{Y}}{f_{X}} = 2 $$.

If the periodic time of a body \( (\mathrm{X}) \) is double that of a body \( (\mathrm{Y}) \) so the ratio between th equency (Y) to the frequency \( (\mathrm{X}) \) is \( \ldots \ldots \)

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