A person is observing two trains one coming towards him and other leaving with the same speed $4\, m/s$. If their whistling frequencies are $240\, Hz$ each, then the number of beats per second heard by the person will be:(if velocity of sound is $320\, m/s$)
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While approaching
$\mathrm{n}^{\prime}=\mathrm{n}\left[\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right]$
while leaving
$\mathrm{n}^{\prime \prime}=\mathrm{n}\left[\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right]$
$\Delta \mathrm{n}=\mathrm{n}^{\prime}-\mathrm{n}^{\prime \prime}=\mathrm{nv}\left[\frac{1}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}-\frac{1}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right]$
$=\frac{2 \mathrm{nv}_{\mathrm{s}}}{\mathrm{v}}=\frac{2 \times 240 \times 4}{320}=6$ Beats per sec.
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