
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 \[320\,m{s^{ - 1}}\].
Answer
225.3k+ views
Hint:A sound is a form of energy that is produced when vibration or a disturbance travels in a medium. The particles of the medium pass on the disturbance as a result of which sound is produced.
Formula used:
Expression of speed is,
$v=\dfrac{s}{t}$
Here, $v$ is the speed, $s$ is the distance and $t$ is the time.
Complete step by step solution:
Given that the distance of the sound wave is = 80m
For an echo to not disturb the listener during the meeting, the sound will be travelling from the building to the loudspeaker and then from the loudspeaker to the building. Therefore, the total distance travelled by the sound wave will be given by
\[s = 80 \times 2 = 160\,m\]
Given that the speed of sound in the air is \[v = 320\,m{s^{ - 1}}\]. Since, the waves are moving in a direction, so the time interval can be calculated using velocity time formula which is,
\[v = \dfrac{s}{t}\]
Or the above equation can also be written as
\[t = \dfrac{s}{v}\]
Substituting all the values given in the question and solving, we get
\[t = \dfrac{{160}}{{320}}\]
\[\Rightarrow t = \dfrac{1}{5}\]
$\therefore t=0.5\,sec$
Therefore, the 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 will be 0.5 seconds.
Note: It is to be remembered that since the particles of the medium are vibrating in the medium, due to those vibrations, regions of high pressure and low pressure will be created. These regions are known as compressions and rarefactions. Due to these being transported to the nearest medium, the sound is produced. The sound produced will be different for different sources which depend on certain characteristics such as frequency, wavelength etc.
Formula used:
Expression of speed is,
$v=\dfrac{s}{t}$
Here, $v$ is the speed, $s$ is the distance and $t$ is the time.
Complete step by step solution:
Given that the distance of the sound wave is = 80m
For an echo to not disturb the listener during the meeting, the sound will be travelling from the building to the loudspeaker and then from the loudspeaker to the building. Therefore, the total distance travelled by the sound wave will be given by
\[s = 80 \times 2 = 160\,m\]
Given that the speed of sound in the air is \[v = 320\,m{s^{ - 1}}\]. Since, the waves are moving in a direction, so the time interval can be calculated using velocity time formula which is,
\[v = \dfrac{s}{t}\]
Or the above equation can also be written as
\[t = \dfrac{s}{v}\]
Substituting all the values given in the question and solving, we get
\[t = \dfrac{{160}}{{320}}\]
\[\Rightarrow t = \dfrac{1}{5}\]
$\therefore t=0.5\,sec$
Therefore, the 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 will be 0.5 seconds.
Note: It is to be remembered that since the particles of the medium are vibrating in the medium, due to those vibrations, regions of high pressure and low pressure will be created. These regions are known as compressions and rarefactions. Due to these being transported to the nearest medium, the sound is produced. The sound produced will be different for different sources which depend on certain characteristics such as frequency, wavelength etc.
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