At a prayer meeting, the disciples sing JAI-RAM JALRAM. The sound amplified by a loudspeaker comes back after reflection from a building at a distance of 80 m from the meeting. What maximum time interval can be kept between one JAI-RAM and the next JAI-RAM so that the echo does not disturb a listener sitting in the meeting. Speed of sound in air is $320 m s^{- 1}$ .

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Solution

Here given, $s = 80 m \times 2 = 160 m$

v =320 m/s

So, the maximum time interval will be

$t = \frac{s}{v} = \frac{160}{320} = 0.5$ seconds

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Here given, s=80m×2=160m v =320 m/s So, the maximum time interval will be t=sv=160320=0.5 seconds

Here given, $s = 80 m \times 2 = 160 m$

v =320 m/s

So, the maximum time interval will be

$t = \frac{s}{v} = \frac{160}{320} = 0.5$ seconds