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.

Open in App

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, $v = \frac{S}{t}$ .
$∴ t = \frac{s}{v} = \frac{160}{320} = 0.5 s$
Therefore, the maximum time interval will be 0.5 seconds.

Suggest Corrections

Similar questions

Shyam is on top of a building of 80m height waving at his friend Ram who is 60m away from the building. what is the direct distance between Ram and Shyam?

Q. Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0 m from one of the speakers and 6.4 m from the other. If the sound signal is continuously varied from 500 Hz to 5000 Hz, what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air = 320 m s⁻¹.

Q. Ram founded a village named Ramsthal. Meena is best friend to Parvathi who is the grand-daughter of Ram. Jai and Preet are the sons of Meena and Parvathi respectively. Also, Karan and Jai are best friends. Parth, who is the current Patla of Ramsthal has fallen ill and wants a new Patla to be chosen. Who among the following is a suitable person to be chosen as the Patla as per the tribal tradition?

Q. Shyam is standing between Ram and a wall on the same line. Distance between Ram and Shyam is

165 m

. Shyam is standing

d

meters away from the wall. Ram claps and Shyam hears the clap two times at an interval of

2

seconds. Find the value of

d

.

(Assume velocity of sound to be

330 m / s

)

Q. The sound emitted by a source returns back, after reflection, from a surface after a time interval of

0.5

seconds. What is the distance between the source and the reflecting surface? (Speed of sound in air is

340 m/s

)

Join BYJU'S Learning Program

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=St. ∴ t=sv=160320=0.5 s Therefore, the maximum time interval will be 0.5 seconds.