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Question
At a prayer meeting, the disciples sing JAI-RAM JAI-RAM. 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
Given:
The distance of the building from the meeting is 80 m.
Velocity of sound in air v = 320 ms−1
Total distance travelled by the sound after echo is S = 80 × 2 = 160 m
As we know,
\[v = \frac{S}{t}\]
\[\therefore t = \frac{s}{v} = \frac{160}{320} = 0 . 5 \text{ s }\]
Therefore, the maximum time interval will be 0.5 seconds.
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