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A man is watching two trains, one leaving and the other coming with equal speed of $4\,m/s$ . If they sound their whistles each of frequency $240\, Hz$ , the number of beats per sec heard by man will be equal to: (velocity of sound in air $= 320\, m/s$ )
$12$
$0$
$3$
$6$
Solution
$\mathrm{n}^{\prime}=\mathrm{n}\left[\frac{\mathrm{v}}{\mathrm{v}-v_{\mathrm{s}}}\right]$
$\mathrm{n}^{\prime \prime}=\mathrm{n}\left[\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right]$
$\Delta \mathrm{n}=\mathrm{n}^{\prime}-\mathrm{n}^{\prime \prime}$
$ = {\rm{nv}}\left[ {\frac{1}{{v – {v_{\rm{s}}}}} – \frac{1}{{v + {v_{\rm{s}}}}}} \right] = {\rm{nv}}\left[ {\frac{{2{{\rm{v}}_{\rm{s}}}}}{{{v^2} – {\rm{v}}_s^2}}} \right]$
$\quad=\frac{2 \mathrm{nv}_{\mathrm{s}}}{\mathrm{v}}=\frac{2 \times 240 \times 4}{320}=6$
